An aluminium piece of mass 50 ,g initially at | vedclass

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Question

(C) According to the principle of calorimetry,heat lost by the aluminium piece is equal to the heat gained by the water.Let $m_a = 50 \,g = 0.05 \,kg$

Vedclass · Accepted Answer

$31.5$

Vedclass · Answer

(C) According to the principle of calorimetry,heat lost by the aluminium piece is equal to the heat gained by the water.Let $m_a = 50 \,g = 0.05 \,kg$ be the mass of aluminium,$s_a = 900 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_a = (300^{\circ} C - 160^{\circ} C) = 140^{\circ} C$ be the change in temperature.Let $m_w = 1 \,kg$ be the mass of water,$s_w = 4200 \,J \,kg^{-1} K^{-1}$ be its specific heat,and $\Delta T_w = (T - 30^{\circ} C)$ be the change in temperature of water.Heat lost by aluminium = $m_a s_a \Delta T_a = 0.05 	imes 900 	imes 140 = 6300 \,J$.Heat gained by water = $m_w s_w \Delta T_w = 1 	imes 4200 	imes (T - 30) = 4200(T - 30)$.Equating the two: $6300 = 4200(T - 30)$.$T - 30 = \frac{6300}{4200} = 1.5$.$T = 30 + 1.5 = 31.5^{\circ} C$.Thus,the final temperature of the water is $31.5^{\circ} C$.

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