# Heat capacity

Physical size
Surname Heat capacity
Formula symbol ${\ displaystyle C}$ Size and
unit system
unit dimension
SI J · K −1 L 2 · M · T −2 · Θ −1

The heat capacity of a body is the ratio of the heat supplied to it ( ) to the resulting temperature increase ( ): ${\ displaystyle C}$ ${\ displaystyle \ Delta Q}$ ${\ displaystyle \ Delta T}$ ${\ displaystyle C = {\ frac {\ mathrm {d} Q} {\ mathrm {d} T}}}$ The unit of heat capacity is J / K .

In the case of homogeneous bodies, the heat capacity can be calculated as the product of the specific heat capacity and the mass of the body, ${\ displaystyle c}$ ${\ displaystyle m}$ ${\ displaystyle C = c \ cdot m,}$ or as a product of its molar heat capacity and its amount of substance : ${\ displaystyle C _ {\ mathrm {m}}}$ ${\ displaystyle n}$ ${\ displaystyle C = C _ {\ mathrm {m}} \ cdot n}$ Both the specific and the molar heat capacity are material constants and are tabulated in relevant reference works.

The heat capacity is an extensive state variable , so it can be calculated for a body that is composed of parts as the sum of the respective heat capacities of its parts. The following therefore results for the total heat capacity : ${\ displaystyle C_ {n}}$ ${\ displaystyle N}$ ${\ displaystyle C _ {\ mathrm {ges}}}$ ${\ displaystyle C _ {\ mathrm {ges}} = \ sum _ {n = 1} ^ {N} C_ {n} = C_ {1} + C_ {2} + \ dotsb + C_ {N}}$ For layer systems such as B. wall constructions , the heat capacity per unit area is given in J / (m 2 · K), for yard goods such as. B. extruded heat sinks per unit length , in J / (m · K).

## Determination of the heat capacity in the mixing test

The experimental determination of the heat capacity of a body shows how to deal with this quantity:

The body is first placed in boiling water ( ) until it has reached this temperature itself. Then you transfer it to a calorimeter in which there is water at the temperature of . A mixing temperature of . ${\ displaystyle \ vartheta _ {1} = 100 \, \ mathrm {^ {\ circ} C}}$ ${\ displaystyle m _ {\ mathrm {W}} = 1 \, \ mathrm {kg}}$ ${\ displaystyle \ vartheta _ {2} = 20 \, \ mathrm {^ {\ circ} C}}$ ${\ displaystyle \ vartheta _ {3} = 30 \, \ mathrm {^ {\ circ} C}}$ So the water has warmed up. ${\ displaystyle \ Delta T _ {\ mathrm {W}} = \ vartheta _ {3} - \ vartheta _ {2} = 10 \, \ mathrm {K}}$ With the known specific heat capacity of water ( ), the heat absorbed by the water is calculated ${\ displaystyle c _ {\ mathrm {W}} \ approx 4 {,} 2 \, \ mathrm {\ tfrac {kJ} {kg \ cdot K}}}$ ${\ displaystyle Q _ {\ mathrm {W}} = c _ {\ mathrm {W}} \ cdot m _ {\ mathrm {W}} \ cdot \ Delta T _ {\ mathrm {W}} = 42 \, \ mathrm {kJ }}$ .

The body has given off this amount of heat to the water when it cools down , so it is . Hence the heat capacity of the body is: ${\ displaystyle \ Delta T _ {\ mathrm {K}} = \ vartheta _ {1} - \ vartheta _ {3} = 70 \, \ mathrm {K}}$ ${\ displaystyle Q _ {\ mathrm {K}} = Q _ {\ mathrm {W}} = 42 \, \ mathrm {kJ}}$ ${\ displaystyle C _ {\ mathrm {K}} = {\ frac {Q _ {\ mathrm {K}}} {\ Delta T _ {\ mathrm {K}}}} = \ mathrm {600 \, {\ frac {J } {K}}}}$ 