A vessel contains 110 , g of water. The heat | vedclass

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(D) Let the final equilibrium temperature be $	heta$.According to the principle of calorimetry,Heat gained by the cold system = Heat lost by the hot

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let the final equilibrium temperature be $	heta$.According to the principle of calorimetry,Heat gained by the cold system = Heat lost by the hot water.The cold system consists of $110 \, g$ of water and the vessel (which has a water equivalent of $10 \, g$).Total mass of cold system = $110 \, g + 10 \, g = 120 \, g$.Heat gained = $m_{cold} \cdot c \cdot (	heta - 	heta_{initial}) = 120 	imes 1 	imes (	heta - 10)$.Heat lost by hot water = $m_{hot} \cdot c \cdot (	heta_{hot} - 	heta) = 220 	imes 1 	imes (70 - 	heta)$.Equating the two: $120(	heta - 10) = 220(70 - 	heta)$.$120	heta - 1200 = 15400 - 220	heta$.$340	heta = 16600$.$	heta = \frac{1660}{34} \approx 48.82^{\circ}C$.Rounding to the nearest provided option,the final temperature is $50^{\circ}C$.

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