Two identical blocks of metal are at 20^{circ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) When two blocks are brought into contact,the hotter block loses heat and the colder block gains heat.Let $T_f$ be the final equilibrium temperatur

Vedclass · Accepted Answer

$T_f$ will be more than $50^{\circ} C$

Vedclass · Answer

(B) When two blocks are brought into contact,the hotter block loses heat and the colder block gains heat.Let $T_f$ be the final equilibrium temperature.According to the principle of calorimetry,heat lost by the hotter block equals heat gained by the colder block.$m \int_{T_f}^{80} s(T) dT = m \int_{20}^{T_f} s(T) dT$Since the specific heat $s(T)$ is an increasing function of temperature,the average specific heat of the hotter block (in the range $T_f$ to $80^{\circ} C$) is greater than the average specific heat of the colder block (in the range $20^{\circ} C$ to $T_f$).Let $s_{avg, hot}$ be the average specific heat of the hotter block and $s_{avg, cold}$ be the average specific heat of the colder block. Then $s_{avg, hot} > s_{avg, cold}$.$s_{avg, hot} (80 - T_f) = s_{avg, cold} (T_f - 20)$$\frac{80 - T_f}{T_f - 20} = \frac{s_{avg, cold}}{s_{avg, hot}} $80 - T_f $100 $T_f > 50^{\circ} C$Therefore,the final temperature will be more than $50^{\circ} C$.

Similar Questions

$5 \ kg$ of water at $20^{\circ} C$ is added to $10 \ kg$ of water at $60^{\circ} C$. Neglecting the heat capacity of the vessel and other losses,the resultant temperature will be nearly: (in $^{\circ} C$)

When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module