Physics · Thermodynamics

Heat transfer and heat capacity: Q = mcΔT

Q: How do you calculate the required heat with $Q = m\cdot{}c\cdot{}\Delta{}T$?

Specify the mass and process, then calculate $\Delta T=T_2-T_1$. Using m in kilograms and suitable c in $\frac{J}{(kg\cdot{}K)}$, $mc\Delta T$ gives heat in joules. For 500 g water heated from 20 to 60 °C with specified constant model value $\frac{4182 J}{(kg\cdot{}K)}$, $Q=83640\,\mathrm J$. Reversing $\frac{Q}{mc}$ must return 40 K. Vessel heat capacity and losses are not yet included. The short equation is appropriate without phase change and with nearly constant c; otherwise use a temperature integral, latent heat or additional energy terms. Count Q positively into the system and negatively when heat leaves.

Q: What is the difference between specific heat capacity and heat capacity?

Specific heat c is the mass-based quantity in $\frac{J}{(kg\cdot{}K)}$. Total heat capacity C of a homogeneous body is $C=mc$, in $\frac{J}{K}$. Doubling mass doubles C at equal state and process conditions while leaving c unchanged. For 0.400 kg with $c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}$, $C=360\,\mathrm{J\,K^{-1}}$ and warming by 20 K needs 7200 J. Specific heat depends on more than material name: temperature, pressure, phase and process matter. Gases have different cp and cv. Add C values for several bodies heated together, such as water and its vessel, provided they undergo the same temperature change.

$Q=mc\Delta T$ calculates heat for a temperature change without a phase transition using suitable approximately constant c. $C=mc$ is heat capacity of the chosen amount.

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Formula

Calculate

Q = m \cdot c \cdot Δ T

Q in joules [J], m in kg, c in J/(kg·K), ΔT in kelvin [K]

Variables & units

Variables & units
Symbol	Meaning	Unit
$Q$	Transferred heat; positive into the selected system	J
$m$	Fixed amount of material by mass	kg
$c$	Specific heat appropriate to state and process; liquid water is often approximated as $\frac{4.18 kJ}{(kg\cdot{}K)}$	J/(kg·K)
$ΔT$	Temperature change T₂ − T₁, not an absolute temperature	K

LaTeX

Q = m \cdot c \cdot \Delta T

QUANTA STUDYquanta-study.de

Heat versus temperature change: two lines with slopes 3.1365 and 3.3165 kilojoules per kelvin. Both start at zero; at 34 kelvin total heat exceeds water heat by 6.120 kilojoules. — Original heating balance with constant appropriate heat capacities: water and vessel need 112.761 kJ over 34 K; water alone needs 106.641 kJ. The vessel contributes 6.120 kJ.

Heat energy calculator

Find the heat, mass, specific heat capacity or temperature change, as long as the substance does not melt or boil.

Find

Mass m

Mass of the substance being heated, e.g. 1 kg for a litre of water.

Specific heat capacity c

Material value from the table, e.g. 4.19 kJ/(kg·K) for water or 0.90 kJ/(kg·K) for aluminium.

Temperature change ΔT

Final minus initial temperature, e.g. 100 °C − 20 °C = 80 K; negative when it cools down.

Filled with example values. Type your own and the result updates straight away.

Result: Heat transferred Q

Q = 335.2 kJ

With m = 1 kg, c = 4.19 kJ/(kg·K) and ΔT = 80 K, Q = 335.2 kJ.

Working · rearrange for Q, substitute, evaluate

1. Rearrange for Q
$Q=mc\Delta T$
2. Substitute values (in SI units)
- c = 4.19 kJ/(kg·K) = 4190 J/(kg·K)
$Q=1\cdot 4190\cdot 80=335200$
3. Result
Q = 335.2 kJ

Quick tips

Estimate for water: warming 1 kg by 1 K takes about 4.2 kJ. Bringing a litre from 20 °C to the boil therefore takes just over 330 kJ, a little under 0.1 kWh.
Temperature differences are the same size in K and °C: from 20 °C to 100 °C is 80 K. Only absolute temperatures need the extra 273.15.

Assumptions and limits

Q = mcΔT for a substance with positive mass and constant process-specific c, with no phase change. Q > 0 means heat input; ΔT = Tfinal − Tinitial. Vessel heat is added separately if needed. ΔT is a temperature difference, including when entered in °C.

Derivation & background

$C=mc$ has unit $\frac{J}{K}$ and c has $\frac{J}{(kg\cdot{}K)}$ . For varying c, $Q=m\int c(T)\,\mathrm dT$ along the specified process. Water’s large c allows much energy for a small temperature change; larger amounts have larger C. Gases require different c values at constant volume and constant pressure. Heat transfer differs from internal energy; pressure–volume or other work must be included in a suitable balance.

Exam blueprint

Validity range

Fixed homogeneous mass without phase change, using specific heat appropriate to process and state; constant c is an interval approximation. Distinguish $\frac{cp}{cv}$ for gases. Vessel heat capacity, losses and other work modes may require additional balances.

Derivation steps

Integrate mass-specific heat capacity along the specified temperature process.

1Specify fixed homogeneous mass, phase and process; heat entering counts positively.
2For small temperature changes, total capacity gives $\delta Q=C\,\mathrm dT$ .
3The mass-specific quantity is $c=\frac Cm$ , in $\mathrm{J\,kg^{-1}\,K^{-1}}$ .
4Substitution gives $\delta Q=mc(T)\,\mathrm dT$ . For gases, c must match constant pressure or volume.
5Integrate over the temperature interval: $Q=m\int_{T_1}^{T_2}c(T)\,\mathrm dT$ .
6For constant c or a suitable interval mean, $Q=mc(T_2-T_1)=mc\Delta T$ . Include vessel and losses separately when relevant.

Rearrangements

Determine specific heat capacity

c = \frac{Q}{m\Delta T}

Temperature differences have the same size in kelvin and °C.

Task variant

Why does water need much energy to heat up?

Water has a high specific heat capacity c.

Common mistakes

Interpreting the sign of ΔT without context.

Heating: Q positive; cooling: Q negative when signs are considered.

Exam context

Typical in calorimetry, mixing temperature and heat exchange.

Formula cluster

Heat theory

Connects energy, temperature and material properties.

η = 1 − T_k/T_w

Carnot Efficiency

pV = nRT

Ideal Gas Law

ΔG = ΔH − TΔS

Gibbs Energy (Free Enthalpy)

P = σ·A·T⁴

Stefan-Boltzmann Law

Worked example

500 g of water is heated from 20 to 60 °C. With specified constant model value $c=4182\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , $Q=mc\Delta T=(0.5)(4182)(40)\,\mathrm J=83640\,\mathrm J\approx83.6\,\mathrm{kJ}$ . Vessel and losses are excluded; checking $\frac{Q}{mc}$ returns 40 K.

Applications

Calorimetry, mixing temperatures and heating/cooling energy requirements; reaction calorimetry with an additional complete energy balance. The equation alone does not determine heat flow rate.

How do you calculate with Heat Capacity: Q = m·c·ΔT?

Here is how to work through a typical Heat Capacity: Q = m·c·ΔT (Q = m·c·ΔT) task step by step:

1
Task
Why does water need much energy to heat up?
Solution path
Water has a high specific heat capacity c.

Understand and apply the equation

$Q=mc\Delta T$ accounts for energy transferred during a temperature change without a phase transition. Here c is material-specific heat capacity and $C=mc$ is the heat capacity of the chosen amount. Process conditions and the system boundary determine whether this calculation captures all supplied energy. Three original examples include vessel heating and a complete calorimeter balance.

Calculating heat: when does Q = m·c·ΔT apply?

For a fixed mass without a phase change, calculate transferred heat as mass times an appropriate approximately constant specific heat capacity times the temperature change. A small metal specimen warms more than a large volume of water when given the same heat. Heat capacity describes the heat required per temperature change for the whole specimen. Specific heat capacity expresses that capacity per unit mass. Material, amount and heating conditions must therefore be kept distinct.[1]

The vessel warms too. If vessel and specimen undergo the same temperature change, their heat capacities add. For mixing, first assign each body its own initial temperature, then determine their common final temperature from the energy balance.[1]

The simple product of mass, specific heat capacity and temperature difference requires a suitable approximately constant material value and no phase change. For a gas, the specification must also say whether volume or pressure is held fixed.[1][3]

Distinguishing heat transfer from stored energy

Q denotes energy transferred across a system boundary because of a temperature difference. A body has internal energy, not a stored state property called heat. Count Q as positive into the body. Heating gives positive Q and cooling negative Q in the simple process considered. Before calculating, decide whether the system includes only water or also the vessel, sensor and supports. The first law of thermodynamics accounts for heat transfer, work and internal-energy change within the same balance.[2]

The equation applies to fixed mass within one phase using heat capacity appropriate to the process. For gases, constant-volume and constant-pressure processes are particularly different. An arbitrary process cannot be represented by an arbitrary tabulated c: pressure–volume work and other work modes can contribute.[3]

From local heat capacity to a finite change

For a specified process, C measures heat needed for a small temperature increase. Divide by mass for a homogeneous amount to obtain $c=\frac Cm$ . Small steps obey $\delta Q=mc(T)\,\mathrm dT$ . The symbol δQ reminds us that heat transfer depends on the path, whereas dT is a temperature increment. Contributions are integrated over the full temperature range.

Q=m\int_{T_1}^{T_2}c(T)\,dT\simeq mc\,(T_2-T_1)=mc\Delta T

The short form needs nearly constant c or an appropriate mean over this particular interval.[1]

\begin{aligned}C&=mc\\\overline c&=\frac1{T_2-T_1}\int_{T_1}^{T_2}c(T)\,\mathrm dT\end{aligned}\quad(T_2\ne T_1)

An interval mean need not equal a table value at one particular temperature.

View diagram: variable heat capacity

Increasing capacity makes each kelvin require more heat

QUANTA STUDYquanta-study.de

For the model specimen in Example 3, integrated heat is

Q=0.400(800\theta+\theta^2)\,\mathrm J

, with

\theta=\Delta T/\mathrm K

. The dashed line uses only the starting value. The increasing slope corresponds to increasing total heat capacity.[1]

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Specific heat capacity c versus total heat capacity C

Specific heat capacity c describes heat required per unit mass and temperature change, in J/(kg·K). Total heat capacity C describes the whole body, in J/K, and for a homogeneous sample equals mass times c. The rearrangements below determine c, ΔT or m from the same heat balance.[1]

\begin{aligned}[c]&=\mathrm{J\,kg^{-1}\,K^{-1}}\\{}[C]&=\mathrm{J\,K^{-1}}\\c&=\frac{Q}{m\Delta T}\\\Delta T&=\frac{Q}{mc}\end{aligned}

Division requires m > 0 and, for the first rearrangement, ΔT ≠ 0.

To find mass, divide the heat absorbed or released by the sample by $c\Delta T$ . The model requires an appropriate approximately constant $c$ for the process, no phase change and $c\Delta T\ne0$ . Use consistent units. For positive $c$ , $Q$ and $\Delta T$ have the same sign, giving positive mass. When $Q=0$ and $\Delta T=0$ , this relation cannot determine mass. Heat for a vessel that also warms must be accounted for separately.[1]

m=\frac{Q}{c\Delta T}\qquad(c\Delta T\ne0)

Mass from the sample’s heat transfer: Q in joules, c in J/(kg·K) and ΔT in kelvin give m in kilograms.[1]

Kelvin and Celsius temperatures differ by 273.15, but their differences have equal numerical values. Going from 18 °C to 52 °C means 34 K, not 325.15 K. Enter 750 g as 0.750 kg. If c were in J/(g·K), grams would be appropriate instead, but do not mix the systems. One kJ/(kg·K) equals 1000 J/(kg·K).[4]

A watt-hour is an energy unit: $1\,\mathrm{Wh}=3600\,\mathrm{J}$ . The value 4182 J/(kg·K) used here therefore equals approximately 1.1617 Wh/(kg·K). A value of 1.163 Wh/(kg·K) instead corresponds to 4186.8 J/(kg·K), approximately 4187 J/(kg·K). Temperature, pressure and process must match the starting value; conversion does not change a material property.[1][7]

Heat storage per unit volume also requires density: $c_{\mathrm{vol}}=\rho c$ . With specified model values of 1000 kg/m³ and 4182 J/(kg·K), the result is 4.182 MJ/(m³·K), or approximately 1.1617 kWh/(m³·K). This is a volumetric heat capacity. It cannot be entered as mass-specific $c$ in the calculator; use the actual mass there.[1][7]

Why material and process determine the value

A large c means more energy for equal mass and temperature change. Water consequently acts as a useful thermal buffer and common coolant. The comparison holds conditions fixed: a large metal object can have greater total C than a small glass of water despite smaller c. Specific heat also depends on temperature, pressure, phase and composition. One water value is not an exact constant for ice, liquid and vapour together.[5]

An ideal gas has cp larger than cv because heating at constant pressure also permits expansion work. On a mass basis, $c_p-c_v=\frac{R}{M_{\mathrm{mol}}}$ , where Mmol is molar mass; the molar difference is R. The first law separates Q, work and internal-energy change. This distinction matters even though all three are measured in joules.[3][2]

Heating time: must the vessel be heated too?

Yes: if the vessel is inside the system boundary and undergoes the same temperature change, include its heat requirement. The heater’s electrical power determines net heating power only after losses are accounted for; the example gives net power directly and adds the heat requirements of the water and vessel.[1][2]

Calculating useful heat and heating time

Given: 0.750 kg of water is heated from 18 °C to 52 °C with constant model value $c_w=4182\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . Vessel capacity $C_{\mathrm{vessel}}=180\,\mathrm{J\,K^{-1}}$ , same initial and final temperatures. Constant net power entering the system is 480 W; no evaporation.
Find: Find total useful heat and ideal heating time.

Approach: water and vessel undergo the same temperature change, so their capacities add. The given net power already enters this system; another efficiency factor would count the loss twice.
$\begin{aligned}C_{\mathrm{tot}}&=m_{\mathrm w}c_{\mathrm w}+C_{\mathrm{vessel}}\\Q&=C_{\mathrm{tot}}\Delta T\\t&=\frac{Q}{P_{\mathrm{net}}}\end{aligned}$
Account for water and vessel separately.
$\begin{aligned}Q_w&=0.750\cdot4182\cdot34=106641\,\mathrm J\\{} Q_v&=180\cdot34=6120\,\mathrm J\end{aligned}$
Add energies and divide by net power.
$\begin{aligned}Q_{\mathrm{total}}&=112761\,\mathrm J\\{} t&=\frac{112761}{480}=234.91875\,\mathrm s\end{aligned}$

Result: About 112.8 kJ and 235 s, or 3.92 min, are required.

Check: Total capacity is $C_{\mathrm{tot}}=(0.750\cdot4182+180)\,\mathrm{J\,K^{-1}}=3316.5\,\mathrm{J\,K^{-1}}$ . Back-calculating $\frac{Q}{C_{\mathrm{tot}}}$ recovers 34 K. Ignoring the vessel misses 6.12 kJ. Net power is the power actually entering, not an unchecked electrical rating.

A mixing temperature with a clear boundary

\begin{aligned}\sum_iC_i(T_f-T_i)&=0\\T_f&=\frac{\sum_iC_iT_i}{\sum_iC_i}\end{aligned}

Closed thermally isolated system without reaction, phase change or significant external work; Ci are constant appropriate heat capacities here.[1]

The warm component has negative Q and the cold one positive Q. Positive heat capacities place Tf between initial temperatures. That check does not apply unchanged to reactions or phase changes, which add energy terms. A calorimeter has its own initial temperature and belongs in the balance as another component.

How do you calculate the equilibrium temperature in a calorimeter?

Finding final temperature with a vessel contribution

Given: A 0.120 kg body with constant model value $c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is at 95 °C. Water: 0.250 kg at 20 °C, $c_w=4182\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . Calorimeter capacity: 90 J/K, also at 20 °C. No losses, reactions or phase changes.
Find: Find Tf and heat released by the hot body.

Approach: In the isolated model, released and absorbed heat sum to zero. $C_h(T_f-T_h)+C_c(T_f-T_c)=0$ . Combine warm and cold heat capacities.
$\begin{aligned}C_h&=0.120\cdot900=108\,\mathrm{J/K}\\{} C_c&=0.250\cdot4182+90=1135.5\,\mathrm{J/K}\end{aligned}$
Solve the signed balance.
$\begin{aligned}108(T_f-95)+1135.5(T_f-20)&=0\\{} T_f&=26.513872\,^\circ\mathrm C\end{aligned}$
Calculate heat for the warm component.
$Q_h=108(26.513872-95)\approx-7396.50\,\mathrm J$

Result: Tf ≈ 26.51 °C; the hot body releases about 7.397 kJ.

Check: The cold side gains $Q_c=(1135.5\,\mathrm{J\,K^{-1}})(6.513872\,\mathrm K)\approx+7396.50\,\mathrm J$ . The sum is zero and Tf lies between 20 and 95 °C; the large cold capacity explains its proximity to 20 °C.

Explore further: variable heat capacity

When capacity varies, integrate the small heat contributions over the temperature interval. A linear model function makes the interval mean easier to understand.[1]

Which heat capacity belongs in the calculation?

Given: A model specimen of $m=0.400\,\mathrm{kg}$ is heated from 300 to 340 K without a phase change. For this process, $c(T)=[800+2(\frac{T}{\mathrm K}-300)]\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is specified. This is a calculation model, not a tabulated material value.
Find: Find the required heat and the appropriate mean value of c.

Approach: integrate the variable material value. For this linear function, the interval mean is the average of its endpoint values.
$\begin{aligned}Q&=m\int_{T_i}^{T_f}c(T)\,\mathrm dT\\\overline c&=\frac{1}{T_f-T_i}\int_{T_i}^{T_f}c(T)\,\mathrm dT\end{aligned}$
Substitute: the starting and final values are 800 and 880 joules per kilogram per kelvin.
$\begin{aligned}\overline c&=\frac{800+880}{2}\,\mathrm{J\,kg^{-1}\,K^{-1}}\\&=840\,\mathrm{J\,kg^{-1}\,K^{-1}}\\Q&=(0.400\,\mathrm{kg})(840\,\mathrm{J\,kg^{-1}\,K^{-1}})(40\,\mathrm K)\\&=13440\,\mathrm J\end{aligned}$

Result: Required heat is 13.44 kJ; the appropriate interval mean is 840 J/(kg·K).

Check: Check: using only the starting value gives 12.80 kJ; using only the final value gives 14.08 kJ. The integral lies between them. This illustrates why a material value should not be assumed constant over a broad temperature interval without checking.

Heat requirement does not determine heat flow rate

$Q=mc\Delta T$ does not say how rapidly heat is transferred. That requires power, conduction or a heat-transfer model. For nonuniform temperature, integrate local energy over mass; one sensor reading is not automatically the whole body’s temperature. Strongly varying c requires the temperature integral. Material values need state and process specifications.

An equilibrium phase change can transfer energy without changing temperature. Then mcΔT gives zero for the transition even though mL is required. Heat-capacity calculations cover temperature segments before and after it. Heat losses, electrical energy and mechanical work additionally belong in a complete device balance.[6][2]

Five solved exercises

Calculate · Foundation

0.200 kg of material with $c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}$ cools from 80 to 25 °C without a phase change. Find heat relative to the body.

Show hint

Q is positive into the body.

Learning goal: Assign sign to cooling heat.

Show solution and calculation

Approach: temperature change is final minus initial temperature.
The suitable constant material value allows mass times specific heat capacity times temperature change.
The negative result means heat rejection: the body loses 5.50 kJ.

Check: $\frac{Q}{mc}$ returns −55 K.

\begin{aligned}\Delta T&=(25-80)\,\mathrm K=-55\,\mathrm K\\Q&=mc\Delta T\\&=(0.200)(500)(-55)\,\mathrm J=-5500\,\mathrm J\end{aligned}

Rearrange · Application

7200 J of heat raises the temperature of 0.400 kg by 20 K with a suitable constant material value. Find specific and total heat capacity.

Show hint

Distinguish total and mass-specific capacities.

Learning goal: Distinguish specific and total heat capacity.

Show solution and calculation

Approach: divide heat by mass and temperature change.
Then multiply specific capacity by mass to find total capacity.

Check: $C\Delta T=7200\,\mathrm J$ recovers the supplied heat. The two capacities have different units.

\begin{aligned}c&=\frac{Q}{m\Delta T}\\&=\frac{7200}{(0.400)(20)}\,\mathrm{J\,kg^{-1}\,K^{-1}}\\&=900\,\mathrm{J\,kg^{-1}\,K^{-1}}\\C&=mc=360\,\mathrm{J\,K^{-1}}\end{aligned}

Calculate · Application

Two bodies mixed in isolation without phase changes have $C_1=200\,\mathrm{J\,K^{-1}}$ at 80 °C and $C_2=600\,\mathrm{J\,K^{-1}}$ at 20 °C. Find their common final temperature.

Show hint

Use the heat-capacity-weighted mean.

Learning goal: Balance isolated mixing temperatures.

Show solution and calculation

Approach: in the isolated system, heat changes sum to zero.
Both bodies share a final temperature but have different temperature changes.
Solve for the shared final temperature.

Check: the hot body loses 9000 J and the cold body gains 9000 J. Final temperature lies between initial readings, closer to the body with the larger capacity.

\begin{aligned}0&=C_1(T_f-T_1)+C_2(T_f-T_2)\\T_f&=\frac{C_1T_1+C_2T_2}{C_1+C_2}\\&=\frac{200\cdot80+600\cdot20}{200+600}\,{}^\circ\mathrm C\\&=35\,{}^\circ\mathrm C\\Q_1&=-9000\,\mathrm J\\Q_2&=+9000\,\mathrm J\end{aligned}

Interpret · Application

Two bodies without phase change each receive the same constant net heating power 10 W, no losses, and have constant capacities. Their linear temperature slopes are 0.10 and 0.25 K/s. Which has larger heat capacity and by what ratio?

Show hint

Greater capacity means smaller slope at equal power.

Learning goal: Relate heating-curve slope to thermal capacity.

Show solution and calculation

Approach: Divide net heating power by each temperature slope.
The slower-heating body has 100 J/K; the other 40 J/K, a ratio of 2.5.
Capacity times slope returns 10 W for both.
Without their masses this does not determine a comparison of specific capacities.

\begin{aligned}P&=C\frac{\mathrm dT}{\mathrm dt}\\{}C_1&=\frac{10}{0.10}\,\mathrm{J\,K^{-1}}=100\,\mathrm{J\,K^{-1}}\\{}C_2&=40\,\mathrm{J\,K^{-1}},&\frac{C_1}{C_2}&=2.5\end{aligned}

[1]

Error analysis · Application

A 0.10 kg specimen with $c=1000\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is in a vessel with $C_{\mathrm{vessel}}=50\,\mathrm{J\,K^{-1}}$ . Both start at 20 °C and together receive 3000 J without losses or phase changes. Someone predicts 50 °C while ignoring the vessel. Find the shared final temperature.

Show hint

Both capacities absorb heat across the same temperature rise.

Learning goal: Do not assign all heating energy solely to the specimen.

Show solution and calculation

Approach: Add specimen capacity 100 J/K and vessel capacity 50 J/K.
The common rise is 20 K and final temperature 40 °C.
The specimen absorbs 2000 J and vessel 1000 J, totaling 3000 J.
The incorrect calculation assigns vessel heat to the specimen and overestimates temperature.

\begin{aligned}C_{\mathrm{tot}}&=mc+C_{\mathrm{vessel}}=150\,\mathrm{J\,K^{-1}}\\{}\Delta T&=\frac{3000\,\mathrm J}{150\,\mathrm{J\,K^{-1}}}=20\,\mathrm K\\{}T_f&=40\,{}^\circ\mathrm C\end{aligned}

[1]

What to carry into your next calculation

For constant $c$ without a phase change, $Q=mc\Delta T$ . Include warmed vessels and integrate when $c(T)$ varies. A temperature difference can be stated in kelvin or Celsius, but an absolute temperature and a temperature difference are different quantities.[1][4]

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

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This explanation, 3 worked examples and 5 exercises are available in German and English. Structure, source references and mathematical notation undergo automated checks. Calculations, model assumptions and both language versions receive a separate AI review.

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An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

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Change history

October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Answered heat-transfer and capacity questions directly, identified vessel and mixing-temperature examples and connected process limits to appropriate follow-up formulas.
October 1, 2026Added the missing rearrangement for mass with process assumptions, sign and unit checks.
October 2, 2026Explained watt-hours and density-dependent volumetric heat capacity while preserving material values.

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Frequently asked questions about Heat Capacity: Q = m·c·ΔT

How do you calculate the required heat with

Q = m\cdot{}c\cdot{}\Delta{}T

Specify the mass and process, then calculate $\Delta T=T_2-T_1$ . Using m in kilograms and suitable c in $\frac{J}{(kg\cdot{}K)}$ , $mc\Delta T$ gives heat in joules. For 500 g water heated from 20 to 60 °C with specified constant model value $\frac{4182 J}{(kg\cdot{}K)}$ , $Q=83640\,\mathrm J$ . Reversing $\frac{Q}{mc}$ must return 40 K. Vessel heat capacity and losses are not yet included. The short equation is appropriate without phase change and with nearly constant c; otherwise use a temperature integral, latent heat or additional energy terms. Count Q positively into the system and negatively when heat leaves.

Why does water need so much energy to heat up?+

Liquid water has a large specific heat under ordinary conditions, roughly $\frac{4.18 kJ}{(kg\cdot{}K)}$ . One kilogram therefore needs about 4.18 kJ for a one-kelvin increase. At equal mass and temperature interval this exceeds many metals’ requirements; do not mix that comparison with differing amounts. A large metal block can have greater total heat capacity than a small glass of water. The property supports thermal storage and cooling, but alone says nothing about heat-transfer speed. The chosen c also belongs to phase, temperature, pressure and process; it is not one universal constant for ice and vapour. Larger mass increases C without changing c at the same conditions.

What is the difference between specific heat capacity and heat capacity?+

Specific heat c is the mass-based quantity in $\frac{J}{(kg\cdot{}K)}$ . Total heat capacity C of a homogeneous body is $C=mc$ , in $\frac{J}{K}$ . Doubling mass doubles C at equal state and process conditions while leaving c unchanged. For 0.400 kg with $c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ , $C=360\,\mathrm{J\,K^{-1}}$ and warming by 20 K needs 7200 J. Specific heat depends on more than material name: temperature, pressure, phase and process matter. Gases have different cp and cv. Add C values for several bodies heated together, such as water and its vessel, provided they undergo the same temperature change.

How do you calculate the mixing temperature of two substances?+

For thermal isolation, constant appropriate heat capacities and no reaction or phase transition, use $C_1(T_f-T_1)+C_2(T_f-T_2)=0$ . This gives $T_f=\frac{C_1T_1+C_2T_2}{C_1+C_2}$ . For $C_1=200\,\mathrm{J\,K^{-1}}$ at 80 °C and $C_2=600\,\mathrm{J\,K^{-1}}$ at 20 °C, the result is 35 °C. Heat contributions −9000 J and +9000 J cancel. A vessel has its own initial temperature and may need another term. With positive capacities Tf lies between initial temperatures. Ice melting, evaporation or an exothermic reaction requires additional energy terms; the simple weighted mean is then insufficient. Checking signs distinguishes heat absorbed from heat rejected, preventing cooling from being interpreted as a positive heating requirement.

Why is ΔT the same in kelvin and Celsius?+

The scales differ by the constant offset 273.15. Subtracting two temperatures cancels it: $(t_2+273.15)-(t_1+273.15)=t_2-t_1$ . Going from 18 to 52 °C therefore means 34 K. Do not add 273.15 again to that difference. This applies to differences, not ratios of absolute temperatures in a gas law or Carnot efficiency. Kelvin in the c unit represents the same-sized temperature increment. Also check subtraction order: cooling from 80 to 25 °C gives $\Delta T=-55\,\mathrm K$ , making Q negative under the chosen convention. Absolute Celsius temperatures still need conversion when an equation requires kelvins.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §1.4 “Heat Transfer, Specific Heat, and Calorimetry”.

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Heat transfer and heat capacity: Q = mcΔT

Q=mc\Delta T

calculates heat for a temperature change without a phase transition using suitable approximately constant c.

C=mc

is heat capacity of the chosen amount.

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Variables & units

Symbol

Meaning

Unit

Q

Transferred heat; positive into the selected system

m

Fixed amount of material by mass

c

Specific heat appropriate to state and process; liquid water is often approximated as

\frac{4.18 kJ}{(kg\cdot{}K)}

J/(kg·K)

ΔT

Temperature change T₂ − T₁, not an absolute temperature

Derivation & background

Exam blueprint

Validity range

Derivation steps

Integrate mass-specific heat capacity along the specified temperature process.

1Specify fixed homogeneous mass, phase and process; heat entering counts positively.
2For small temperature changes, total capacity gives $\delta Q=C\,\mathrm dT$ .
3The mass-specific quantity is $c=\frac Cm$ , in $\mathrm{J\,kg^{-1}\,K^{-1}}$ .
4Substitution gives $\delta Q=mc(T)\,\mathrm dT$ . For gases, c must match constant pressure or volume.
5Integrate over the temperature interval: $Q=m\int_{T_1}^{T_2}c(T)\,\mathrm dT$ .
6For constant c or a suitable interval mean, $Q=mc(T_2-T_1)=mc\Delta T$ . Include vessel and losses separately when relevant.

Understand and apply the equation

Q=mc\Delta T

accounts for energy transferred during a temperature change without a phase transition. Here c is material-specific heat capacity and

C=mc

is the heat capacity of the chosen amount. Process conditions and the system boundary determine whether this calculation captures all supplied energy. Three original examples include vessel heating and a complete calorimeter balance.

Calculating heat: when does Q = m·c·ΔT apply?

Distinguishing heat transfer from stored energy

From local heat capacity to a finite change

Q=m\int_{T_1}^{T_2}c(T)\,dT\simeq mc\,(T_2-T_1)=mc\Delta T

The short form needs nearly constant c or an appropriate mean over this particular interval.[1]

\begin{aligned}C&=mc\\\overline c&=\frac1{T_2-T_1}\int_{T_1}^{T_2}c(T)\,\mathrm dT\end{aligned}\quad(T_2\ne T_1)

An interval mean need not equal a table value at one particular temperature.

View diagram: variable heat capacity

Increasing capacity makes each kelvin require more heat

QUANTA STUDYquanta-study.de

For the model specimen in Example 3, integrated heat is

Q=0.400(800\theta+\theta^2)\,\mathrm J

, with

\theta=\Delta T/\mathrm K

. The dashed line uses only the starting value. The increasing slope corresponds to increasing total heat capacity.[1]

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Specific heat capacity c versus total heat capacity C

\begin{aligned}[c]&=\mathrm{J\,kg^{-1}\,K^{-1}}\\{}[C]&=\mathrm{J\,K^{-1}}\\c&=\frac{Q}{m\Delta T}\\\Delta T&=\frac{Q}{mc}\end{aligned}

Division requires m > 0 and, for the first rearrangement, ΔT ≠ 0.

m=\frac{Q}{c\Delta T}\qquad(c\Delta T\ne0)

Mass from the sample’s heat transfer: Q in joules, c in J/(kg·K) and ΔT in kelvin give m in kilograms.[1]

Why material and process determine the value

Heating time: must the vessel be heated too?

Calculating useful heat and heating time

Given: 0.750 kg of water is heated from 18 °C to 52 °C with constant model value $c_w=4182\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . Vessel capacity $C_{\mathrm{vessel}}=180\,\mathrm{J\,K^{-1}}$ , same initial and final temperatures. Constant net power entering the system is 480 W; no evaporation.
Find: Find total useful heat and ideal heating time.

Approach: water and vessel undergo the same temperature change, so their capacities add. The given net power already enters this system; another efficiency factor would count the loss twice.
$\begin{aligned}C_{\mathrm{tot}}&=m_{\mathrm w}c_{\mathrm w}+C_{\mathrm{vessel}}\\Q&=C_{\mathrm{tot}}\Delta T\\t&=\frac{Q}{P_{\mathrm{net}}}\end{aligned}$
Account for water and vessel separately.
$\begin{aligned}Q_w&=0.750\cdot4182\cdot34=106641\,\mathrm J\\{} Q_v&=180\cdot34=6120\,\mathrm J\end{aligned}$
Add energies and divide by net power.
$\begin{aligned}Q_{\mathrm{total}}&=112761\,\mathrm J\\{} t&=\frac{112761}{480}=234.91875\,\mathrm s\end{aligned}$

Result: About 112.8 kJ and 235 s, or 3.92 min, are required.

A mixing temperature with a clear boundary

\begin{aligned}\sum_iC_i(T_f-T_i)&=0\\T_f&=\frac{\sum_iC_iT_i}{\sum_iC_i}\end{aligned}

Closed thermally isolated system without reaction, phase change or significant external work; Ci are constant appropriate heat capacities here.[1]

How do you calculate the equilibrium temperature in a calorimeter?

Finding final temperature with a vessel contribution

Given: A 0.120 kg body with constant model value $c=900\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is at 95 °C. Water: 0.250 kg at 20 °C, $c_w=4182\,\mathrm{J\,kg^{-1}\,K^{-1}}$ . Calorimeter capacity: 90 J/K, also at 20 °C. No losses, reactions or phase changes.
Find: Find Tf and heat released by the hot body.

Approach: In the isolated model, released and absorbed heat sum to zero. $C_h(T_f-T_h)+C_c(T_f-T_c)=0$ . Combine warm and cold heat capacities.
$\begin{aligned}C_h&=0.120\cdot900=108\,\mathrm{J/K}\\{} C_c&=0.250\cdot4182+90=1135.5\,\mathrm{J/K}\end{aligned}$
Solve the signed balance.
$\begin{aligned}108(T_f-95)+1135.5(T_f-20)&=0\\{} T_f&=26.513872\,^\circ\mathrm C\end{aligned}$
Calculate heat for the warm component.
$Q_h=108(26.513872-95)\approx-7396.50\,\mathrm J$

Result: Tf ≈ 26.51 °C; the hot body releases about 7.397 kJ.

Explore further: variable heat capacity

When capacity varies, integrate the small heat contributions over the temperature interval. A linear model function makes the interval mean easier to understand.[1]

Which heat capacity belongs in the calculation?

Given: A model specimen of $m=0.400\,\mathrm{kg}$ is heated from 300 to 340 K without a phase change. For this process, $c(T)=[800+2(\frac{T}{\mathrm K}-300)]\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is specified. This is a calculation model, not a tabulated material value.
Find: Find the required heat and the appropriate mean value of c.

Approach: integrate the variable material value. For this linear function, the interval mean is the average of its endpoint values.
$\begin{aligned}Q&=m\int_{T_i}^{T_f}c(T)\,\mathrm dT\\\overline c&=\frac{1}{T_f-T_i}\int_{T_i}^{T_f}c(T)\,\mathrm dT\end{aligned}$
Substitute: the starting and final values are 800 and 880 joules per kilogram per kelvin.
$\begin{aligned}\overline c&=\frac{800+880}{2}\,\mathrm{J\,kg^{-1}\,K^{-1}}\\&=840\,\mathrm{J\,kg^{-1}\,K^{-1}}\\Q&=(0.400\,\mathrm{kg})(840\,\mathrm{J\,kg^{-1}\,K^{-1}})(40\,\mathrm K)\\&=13440\,\mathrm J\end{aligned}$

Result: Required heat is 13.44 kJ; the appropriate interval mean is 840 J/(kg·K).

Heat requirement does not determine heat flow rate

Five solved exercises

Calculate · Foundation

0.200 kg of material with $c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}$ cools from 80 to 25 °C without a phase change. Find heat relative to the body.

Show hint

Q is positive into the body.

Learning goal: Assign sign to cooling heat.

Show solution and calculation

Approach: temperature change is final minus initial temperature.
The suitable constant material value allows mass times specific heat capacity times temperature change.
The negative result means heat rejection: the body loses 5.50 kJ.

Check: $\frac{Q}{mc}$ returns −55 K.

\begin{aligned}\Delta T&=(25-80)\,\mathrm K=-55\,\mathrm K\\Q&=mc\Delta T\\&=(0.200)(500)(-55)\,\mathrm J=-5500\,\mathrm J\end{aligned}

Rearrange · Application

7200 J of heat raises the temperature of 0.400 kg by 20 K with a suitable constant material value. Find specific and total heat capacity.

Show hint

Distinguish total and mass-specific capacities.

Learning goal: Distinguish specific and total heat capacity.

Show solution and calculation

Approach: divide heat by mass and temperature change.
Then multiply specific capacity by mass to find total capacity.

Check: $C\Delta T=7200\,\mathrm J$ recovers the supplied heat. The two capacities have different units.

\begin{aligned}c&=\frac{Q}{m\Delta T}\\&=\frac{7200}{(0.400)(20)}\,\mathrm{J\,kg^{-1}\,K^{-1}}\\&=900\,\mathrm{J\,kg^{-1}\,K^{-1}}\\C&=mc=360\,\mathrm{J\,K^{-1}}\end{aligned}

Calculate · Application

Two bodies mixed in isolation without phase changes have $C_1=200\,\mathrm{J\,K^{-1}}$ at 80 °C and $C_2=600\,\mathrm{J\,K^{-1}}$ at 20 °C. Find their common final temperature.

Show hint

Use the heat-capacity-weighted mean.

Learning goal: Balance isolated mixing temperatures.

Show solution and calculation

Approach: in the isolated system, heat changes sum to zero.
Both bodies share a final temperature but have different temperature changes.
Solve for the shared final temperature.

Check: the hot body loses 9000 J and the cold body gains 9000 J. Final temperature lies between initial readings, closer to the body with the larger capacity.

\begin{aligned}0&=C_1(T_f-T_1)+C_2(T_f-T_2)\\T_f&=\frac{C_1T_1+C_2T_2}{C_1+C_2}\\&=\frac{200\cdot80+600\cdot20}{200+600}\,{}^\circ\mathrm C\\&=35\,{}^\circ\mathrm C\\Q_1&=-9000\,\mathrm J\\Q_2&=+9000\,\mathrm J\end{aligned}

Interpret · Application

Two bodies without phase change each receive the same constant net heating power 10 W, no losses, and have constant capacities. Their linear temperature slopes are 0.10 and 0.25 K/s. Which has larger heat capacity and by what ratio?

Show hint

Greater capacity means smaller slope at equal power.

Learning goal: Relate heating-curve slope to thermal capacity.

Show solution and calculation

Approach: Divide net heating power by each temperature slope.
The slower-heating body has 100 J/K; the other 40 J/K, a ratio of 2.5.
Capacity times slope returns 10 W for both.
Without their masses this does not determine a comparison of specific capacities.

\begin{aligned}P&=C\frac{\mathrm dT}{\mathrm dt}\\{}C_1&=\frac{10}{0.10}\,\mathrm{J\,K^{-1}}=100\,\mathrm{J\,K^{-1}}\\{}C_2&=40\,\mathrm{J\,K^{-1}},&\frac{C_1}{C_2}&=2.5\end{aligned}

[1]

Error analysis · Application

A 0.10 kg specimen with $c=1000\,\mathrm{J\,kg^{-1}\,K^{-1}}$ is in a vessel with $C_{\mathrm{vessel}}=50\,\mathrm{J\,K^{-1}}$ . Both start at 20 °C and together receive 3000 J without losses or phase changes. Someone predicts 50 °C while ignoring the vessel. Find the shared final temperature.

Show hint

Both capacities absorb heat across the same temperature rise.

Learning goal: Do not assign all heating energy solely to the specimen.

Show solution and calculation

Approach: Add specimen capacity 100 J/K and vessel capacity 50 J/K.
The common rise is 20 K and final temperature 40 °C.
The specimen absorbs 2000 J and vessel 1000 J, totaling 3000 J.
The incorrect calculation assigns vessel heat to the specimen and overestimates temperature.

\begin{aligned}C_{\mathrm{tot}}&=mc+C_{\mathrm{vessel}}=150\,\mathrm{J\,K^{-1}}\\{}\Delta T&=\frac{3000\,\mathrm J}{150\,\mathrm{J\,K^{-1}}}=20\,\mathrm K\\{}T_f&=40\,{}^\circ\mathrm C\end{aligned}

[1]

What to carry into your next calculation

Sources for this explanation

Quanta · Updated October 2, 2026

Editorial responsibility: Quanta Study editorial team

Review scope

An internal human subject review is documented for an earlier release of this page. That confirmation does not apply automatically to the revised explanation and new exercises.

An independent, named human subject review has not yet been documented. Our exercise numbers are worked examples, rather than measurements published by the cited sources.

Editorial responsibility, sources and methods

Change history

October 1, 2026Expanded explanations and learning transitions; added a further fully worked example and an original quantitative diagram; clarified assumptions and calculation steps.
October 1, 2026Added two exercises; labelled all five by task type, difficulty and learning goal.
October 1, 2026Answered heat-transfer and capacity questions directly, identified vessel and mixing-temperature examples and connected process limits to appropriate follow-up formulas.
October 1, 2026Added the missing rearrangement for mass with process assumptions, sign and unit checks.
October 2, 2026Explained watt-hours and density-dependent volumetric heat capacity while preserving material values.

Cite this reference page

Quanta. Heat Capacity: Q = m·c·ΔT. Updated October 2, 2026. https://quanta-study.de/en/formula-sheet/waermekapazitaet

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Frequently asked questions about Heat Capacity: Q = m·c·ΔT

How do you calculate the required heat with

Q = m\cdot{}c\cdot{}\Delta{}T

Why does water need so much energy to heat up?+

What is the difference between specific heat capacity and heat capacity?+

How do you calculate the mixing temperature of two substances?+

Why is ΔT the same in kelvin and Celsius?+

Heat transfer and heat capacity: Q = mcΔT

Variables & units

Heat energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Heat theory

Worked example

Applications

How do you calculate with Heat Capacity: Q = m·c·ΔT?

Calculating heat: when does Q = m·c·ΔT apply?

Distinguishing heat transfer from stored energy

From local heat capacity to a finite change

Increasing capacity makes each kelvin require more heat

Specific heat capacity c versus total heat capacity C

Why material and process determine the value

Heating time: must the vessel be heated too?

Calculating useful heat and heating time

A mixing temperature with a clear boundary

How do you calculate the equilibrium temperature in a calorimeter?

Finding final temperature with a vessel contribution

Explore further: variable heat capacity

Which heat capacity belongs in the calculation?

Heat requirement does not determine heat flow rate

Five solved exercises

0.200 kg of material with c=500 J kg−1 K−1c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}c=500Jkg−1K−1 cools from 80 to 25 °C without a phase change. Find heat relative to the body.

7200 J of heat raises the temperature of 0.400 kg by 20 K with a suitable constant material value. Find specific and total heat capacity.

Two bodies mixed in isolation without phase changes have C1=200 J K−1C_1=200\,\mathrm{J\,K^{-1}}C1​=200JK−1 at 80 °C and C2=600 J K−1C_2=600\,\mathrm{J\,K^{-1}}C2​=600JK−1 at 20 °C. Find their common final temperature.

Two bodies without phase change each receive the same constant net heating power 10 W, no losses, and have constant capacities. Their linear temperature slopes are 0.10 and 0.25 K/s. Which has larger heat capacity and by what ratio?

What to carry into your next calculation

Sources for this explanation

Frequently asked questions about Heat Capacity: Q = m·c·ΔT

Subject references and further reading

Related formulas

More formulas: Heat and thermodynamics

Retain Heat Capacity: Q = m·c·ΔT for exams

Heat transfer and heat capacity: Q = mcΔT

Variables & units

Heat energy calculator

Quick tips

Derivation & background

Exam blueprint

Validity range

Derivation steps

Rearrangements

Task variant

Common mistakes

Exam context

Heat theory

Worked example

Applications

How do you calculate with Heat Capacity: Q = m·c·ΔT?

Calculating heat: when does Q = m·c·ΔT apply?

Distinguishing heat transfer from stored energy

From local heat capacity to a finite change

Increasing capacity makes each kelvin require more heat

Specific heat capacity c versus total heat capacity C

Why material and process determine the value

Heating time: must the vessel be heated too?

Calculating useful heat and heating time

A mixing temperature with a clear boundary

How do you calculate the equilibrium temperature in a calorimeter?

Finding final temperature with a vessel contribution

Explore further: variable heat capacity

Which heat capacity belongs in the calculation?

Heat requirement does not determine heat flow rate

Five solved exercises

0.200 kg of material with c=500 J kg−1 K−1c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}c=500Jkg−1K−1 cools from 80 to 25 °C without a phase change. Find heat relative to the body.

7200 J of heat raises the temperature of 0.400 kg by 20 K with a suitable constant material value. Find specific and total heat capacity.

Two bodies mixed in isolation without phase changes have C1=200 J K−1C_1=200\,\mathrm{J\,K^{-1}}C1​=200JK−1 at 80 °C and C2=600 J K−1C_2=600\,\mathrm{J\,K^{-1}}C2​=600JK−1 at 20 °C. Find their common final temperature.

Two bodies without phase change each receive the same constant net heating power 10 W, no losses, and have constant capacities. Their linear temperature slopes are 0.10 and 0.25 K/s. Which has larger heat capacity and by what ratio?

What to carry into your next calculation

Sources for this explanation

Frequently asked questions about Heat Capacity: Q = m·c·ΔT

0.200 kg of material with $c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}$ cools from 80 to 25 °C without a phase change. Find heat relative to the body.

Two bodies mixed in isolation without phase changes have $C_1=200\,\mathrm{J\,K^{-1}}$ at 80 °C and $C_2=600\,\mathrm{J\,K^{-1}}$ at 20 °C. Find their common final temperature.

0.200 kg of material with $c=500\,\mathrm{J\,kg^{-1}\,K^{-1}}$ cools from 80 to 25 °C without a phase change. Find heat relative to the body.

Two bodies mixed in isolation without phase changes have $C_1=200\,\mathrm{J\,K^{-1}}$ at 80 °C and $C_2=600\,\mathrm{J\,K^{-1}}$ at 20 °C. Find their common final temperature.