An unknown metal of mass 192 , g heated to a | vedclass

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(B) According to the principle of calorimetry, $Heat \, lost \, by \, metal = Heat \, gained \, by \, calorimeter + Heat \, gained \, by \, water$.Let

Vedclass · Accepted Answer

$920$

Vedclass · Answer

(B) According to the principle of calorimetry, $Heat \, lost \, by \, metal = Heat \, gained \, by \, calorimeter + Heat \, gained \, by \, water$.Let $S$ be the specific heat of the metal in $J \, kg^{-1} \, K^{-1}$.Mass of metal $m_m = 0.192 \, kg$, initial temperature $T_m = 100 \, ^oC$, final temperature $T_f = 21.5 \, ^oC$.Mass of calorimeter $m_c = 0.128 \, kg$, specific heat $c_c = 394 \, J \, kg^{-1} \, K^{-1}$, initial temperature $T_i = 8.4 \, ^oC$.Mass of water $m_w = 0.240 \, kg$, specific heat $c_w = 4186 \, J \, kg^{-1} \, K^{-1}$.$0.192 	imes S 	imes (100 - 21.5) = (0.128 	imes 394 + 0.240 	imes 4186) 	imes (21.5 - 8.4)$$0.192 	imes S 	imes 78.5 = (50.432 + 1004.64) 	imes 13.1$$15.072 	imes S = 1055.072 	imes 13.1$$15.072 	imes S = 13821.4432$$S \approx 917 \, J \, kg^{-1} \, K^{-1}$.Rounding to the nearest provided option, the answer is $920 \, J \, kg^{-1} \, K^{-1}$.

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