Heat capacity

This article is about a body's heat capacity. For material properties, see specific 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: ${\displaystyle \Delta T_{\mathrm {K} }=\vartheta _{1}-\vartheta _{3}=70\,\mathrm {K} }$ ${\displaystyle Q_{\mathrm {K} }=Q_{\mathrm {W} }=42\,\mathrm {kJ} }$

${\displaystyle C_{\mathrm {K} }={\frac {Q_{\mathrm {K} }}{\Delta T_{\mathrm {K} }}}=\mathrm {600\,{\frac {J}{K}}} }$