Heat Capacity: Q = m·c·ΔT

Multiply the mass m in kilograms, the specific heat capacity c in J/(kg·K) and the temperature change ΔT in kelvin: Q = m·c·ΔT. The result is the heat added or removed in joules. Example: 500 g of water with c = 4182 J/(kg·K), heated from 20 °C to 60 °C, needs Q = 0.5·4182·40 = 83 640 J ≈ 83.6 kJ. Make sure to insert the mass in kilograms; 500 g is 0.5 kg. The temperature difference ΔT is the same in kelvin and Celsius, because both scales have the same step size. For ΔT only the difference matters, not the absolute value.

Water has an unusually high specific heat capacity of about 4182 J/(kg·K), much higher than most substances; metals often lie at only a few hundred J/(kg·K). This means you must supply a lot of energy to heat one kilogram of water by just one kelvin. The cause is the strong hydrogen bonds between the molecules, which absorb a large part of the energy. This property has major consequences: oceans store enormous amounts of heat and buffer the climate, the body regulates its temperature through water, and water is suitable as a coolant. Conversely water releases correspondingly much energy when cooling down.

The specific heat capacity c is a material property and gives the energy needed to heat one kilogram of a substance by one kelvin, in J/(kg·K). It is independent of the amount and depends only on the material. The heat capacity C, by contrast, refers to a specific body and is the product of mass and specific heat capacity, C = m·c, in J/K. It states how much energy this particular body needs to warm up by one kelvin. A large water tank has a high heat capacity, although its specific heat capacity is the same as that of a glass of water. Do not confuse the material-based c with the body-based C.

For a mixture you use energy conservation: the heat the warmer body releases is taken up by the colder one. You set the released and absorbed heat equal, each calculated with Q = m·c·ΔT. For two substances m₁·c₁·(T_mix − T₁) + m₂·c₂·(T_mix − T₂) = 0 holds, if no heat is lost to the surroundings. Solving for the mixing temperature gives a weighted average of the initial temperatures, where the products m·c are the weights. A substance with large mass or high heat capacity determines the final temperature more strongly. This calorimetry calculation is a classic problem type; watch for consistent units and the sign of the temperature differences.

Because the Kelvin and Celsius scales have the same step size; they are only shifted by 273.15 against each other. A warming of 40 degrees Celsius corresponds exactly to a warming of 40 kelvin, because in forming the difference the constant shift cancels out. Therefore in Q = m·c·ΔT you may conveniently compute the temperature difference in Celsius, and the result is identical to the one in kelvin. This distinction only matters for absolute temperatures, for example in the ideal gas law or the Carnot efficiency, where the absolute value and not the difference counts. For heat capacity, by contrast, you always work with a difference, so both scales are equivalent.