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This is a problem I came across: Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at 100 ºC, while the other one is at 0 ºC. If the two bodies are brought into contact, then assuming no heat loss, the final common temperature is (A) 50 degree Celsius (B) less than 50 degrees but greater than 0 degree Celsius (C) more than 50 degrees (D) none

The answer is (C) more than 50 degrees Celsius. I do not understand why. No relation of temperature dependence with heat capacity is given, whether it varies linearly or in another power of temperature,etc.

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  • $\begingroup$ If there is an increase in specific heat capacity with temperature then the "average" specific heat capacity of the body which stays at the higher temperature until equilibrium is reached will be larger than the "average" specific heat capacity of the other body. $\endgroup$ – Farcher Mar 7 '18 at 8:52
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I think that a possible and very simple argument may be just the following.

The heat capacity is the quantity of heat needed to vary the temperature of a certain amount. If it grows with the temperature, no matter what the exact dependence, than we know that higher the temperature, the greater the heat needed to vary it (increase or decrease). Hence, for an equal amount of heat exchanged between the two bodies (absorbed by the cold one and release by the other), the variation of the temperature must larger for the the cold body with respect to the hot body, since the exchange of heat stops when they reach the same temperature. Therefore, when the equilibrium is reached, the final temperature will be higher with respect to the case of constant heat capacity.

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If the heat capacity varies with temperature, then the heat balance condition for determining the final equilibrium temperature T in this problem is: $$\int_0^T{C(T')dT'}=\int_T^{100}{C(T')dT'} $$where C(T') is the heat capacity at temperature T'. From the mean value theorem for integration, we know that $$\int_0^T{C(T')dT'}=C(T_1)T$$and $$\int_T^{100}{C(T')dT'}=C(T_2)(T-100)$$where $T_1$ is some temperature between 0 and T, and $T_2$ is some temperature between T and 100. So, we have: $$C(T_1)T=C(T_2)(T-100)$$So, solving for T, we obtain: $$T=100\frac{C(T_2)}{(C(T_1)+C(T_2)}=50\left[1+\frac{C(T_2)-C(T_1)}{(C(T_1)+C(T_2))}\right]$$

From the problem statement, we know that the heat capacity increases with temperature, so $C(T_2)>C(T_1)$. This means that the second term in brackets is positive, and the final temperature is greater than 50 C.

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