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With approximately constant specific heat capacity, sensible heat is proportional to mass and temperature change.
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Formula
Q in J; m in kg; cp in J/(kg·K); ΔT in K
| Symbol | Meaning | Unit |
|---|---|---|
| Sensible heat transferred | J | |
| Mass | kg | |
| Specific heat capacity at constant pressure | J/(kg·K) | |
| Temperature change | K |
Q=mc_p\Delta TThis integrates dQ=mc_p(T)dT while approximating cp as constant over the temperature range. It excludes phase change and chemical reaction. Large temperature spans, varying properties or open-flow systems require property integration or an enthalpy balance.
This integrates dQ=mc_p(T)dT while approximating cp as constant over the temperature range. It excludes phase change and chemical reaction. Large temperature spans, varying properties or open-flow systems require property integration or an enthalpy balance.
With approximately constant specific heat capacity, sensible heat is proportional to mass and temperature change.
Base equation for the target quantity
Preserve signs, boundary conditions and units when rearranging.
Work through the numerical example for “Sensible heat at constant heat capacity” and check its unit and magnitude.
1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change.
Substituting a related quantity that shares the same symbol.
Use variable definitions and check: Q in J; m in kg; cp in J/(kg·K); ΔT in K.
Formula cluster
With approximately constant specific heat capacity, sensible heat is proportional to mass and temperature change.
1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change.
Estimate heating or cooling of a material without a phase change.
Here is how to work through a typical Sensible heat at constant heat capacity (Q = m cp ΔT) task step by step:
Task
Work through the numerical example for “Sensible heat at constant heat capacity” and check its unit and magnitude.
Solution path
1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change.
If heat capacity changes little over the interval, a representative value can be used. For wide ranges integrate cp(T) over temperature. Phase changes require separate latent heat; this sensible-heat equation does not include it. Assume no phase change and constant heat capacity only over the stated temperature interval. Application: Estimate heating or cooling of a material without a phase change. The numerical example 1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change anchors the idea in a concrete measurement case.
Temperature differences have the same numerical value in kelvin and Celsius because the scales differ only by an offset. Do not confuse an absolute Celsius temperature with ΔT. A 10 °C rise is ΔT=10 K here. Assume no phase change and constant heat capacity only over the stated temperature interval. The corresponding units are Q in J; m in kg; cp in J/(kg·K); ΔT in K. 1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change. This checks that the reference quantities are compatible.
Kilograms times joules per kilogram-kelvin times kelvin reduces to joules. If cp is entered in kJ/(kg·K), Q is in kJ absent other conversions. This check helps distinguish heat capacity C in J/K from specific heat cp. Assume no phase change and constant heat capacity only over the stated temperature interval. Scope: This integrates dQ=mc_p(T)dT while approximating cp as constant over the temperature range. It excludes phase change and chemical reaction. Large temperature spans, varying properties or open-flow systems require property integration or an enthalpy balance. A practical application is Estimate heating or cooling of a material without a phase change.
Write the quantity and unit from the variable list next to every value: Q in J; m in kg; cp in J/(kg·K); ΔT in K. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example 1 kg of water with cp≈4.18 kJ/(kg·K) needs about 41.8 kJ for a 10 K rise, with no phase change. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. Assume no phase change and constant heat capacity only over the stated temperature interval.
The background states this starting condition: This integrates dQ=mc_p(T)dT while approximating cp as constant over the temperature range. It excludes phase change and chemical. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. Assume no phase change and constant heat capacity only over the stated temperature interval.
Quanta exam set
Curated exam set for "Sensible heat at constant heat capacity":
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Which formula describes Sensible heat at constant heat capacity?
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How do you rearrange Sensible heat at constant heat capacity for Base equation for the target quantity?
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Which common mistake happens with Sensible heat at constant heat capacity?
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