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

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(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 $50^{\circ} C$.

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