200 , g of a solid ball at 20 , ^circ C is dr | vedclass

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(B) According to the principle of calorimetry,heat lost by water equals heat gained by the solid ball.Heat lost by water = $m 	imes S_w 	imes \Delta

Vedclass · Accepted Answer

One half of water

Vedclass · Answer

(B) According to the principle of calorimetry,heat lost by water equals heat gained by the solid ball.Heat lost by water = $m 	imes S_w 	imes \Delta T_w = 200 	imes S_w 	imes (80 - 60) = 200 	imes S_w 	imes 20$.Heat gained by solid = $m 	imes S_s 	imes \Delta T_s = 200 	imes S_s 	imes (60 - 20) = 200 	imes S_s 	imes 40$.Equating the two: $200 	imes S_w 	imes 20 = 200 	imes S_s 	imes 40$.$20 	imes S_w = 40 	imes S_s$.$S_s = \frac{20}{40} 	imes S_w = 0.5 	imes S_w$.Therefore,the specific heat of the solid is one half of that of water.

$200 \, g$ of a solid ball at $20 \, ^\circ C$ is dropped in an equal amount of water at $80 \, ^\circ C$. The resulting temperature is $60 \, ^\circ C$. This means that the specific heat of the solid is

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