A piece of copper of mass 250 g at 500^{circ | vedclass

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(C) According to the principle of calorimetry,Heat lost by the copper = Heat gained by the water and calorimeter.Let $S$ be the specific heat of coppe

Vedclass · Accepted Answer

$380$

Vedclass · Answer

(C) According to the principle of calorimetry,Heat lost by the copper = Heat gained by the water and calorimeter.Let $S$ be the specific heat of copper.Heat lost by copper = $m_{Cu} \cdot S \cdot \Delta T_{Cu} = 0.250 \ kg \cdot S \cdot (500^{\circ} C - 60^{\circ} C) = 0.250 \cdot S \cdot 440 = 110S$.Heat gained by water and calorimeter = $(m_{water} + w) \cdot S_{water} \cdot \Delta T_{water}$,where $w$ is the water equivalent.Heat gained = $(0.200 \ kg + 0.050 \ kg) \cdot 4200 \ J/kg^{\circ} C \cdot (60^{\circ} C - 20^{\circ} C) = 0.250 \cdot 4200 \cdot 40 = 42000 \ J$.Equating the two: $110S = 42000$.$S = 42000 / 110 \approx 381.8 \ J/kg^{\circ} C$.Rounding to the nearest given option,$S \approx 380 \ J/kg^{\circ} C$.

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