Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is at $100^\circ\text{C}$, while the other one is at $0^\circ\text{C}$. If the two bodies are brought into contact, then, assuming no heat loss, the final common temperature is:
(2016 – II)
1. Less than $50^\circ\text{C}$ but greater than $0^\circ\text{C}$
2. $0^\circ\text{C}$
3. $50^\circ\text{C}$
4. More than $50^\circ\text{C}$
View Answer
By conservation of energy, $\int_{T_f}^{100} C(T)dT = \int_{0}^{T_f} C(T)dT$. Since $C(T)$ is greater at higher temperatures, the change in temperature for the hotter body will be less than that for the colder body. Thus, $T_f > 50^\circ\text{C}$.
When $1 text{ kg}$ of ice at $0^circ text{C}$ melts to water at $0^circ text{C}$, the resulting change in its entropy, taking latent heat of ice to be $80 text{ Cal}/^circ text{C}$, is:
(2011 Pre)
1. $273 text{ cal/K}$
2. $8 times 10^4 text{ cal/K}$
3. $80 text{ cal/K}$
4. $293 text{ cal/K}$
View Answer
Entropy change $Delta S = frac{Delta Q}{T} = frac{m cdot L}{T} = frac{1000 cdot 80}{273} = 293 text{ cal/K}$