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

$49$

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.Heat gained by the vessel and the initial water: $(m_{vessel} \cdot c + m_{water1} \cdot c) \cdot (	heta - T_{initial}) = (10 + 110) \cdot 1 \cdot (	heta - 10) = 120(	heta - 10)$.Heat lost by the added hot water: $m_{water2} \cdot c \cdot (T_{hot} - 	heta) = 220 \cdot 1 \cdot (70 - 	heta) = 220(70 - 	heta)$.Equating the two: $120(	heta - 10) = 220(70 - 	heta)$.Dividing by $20$: $6(	heta - 10) = 11(70 - 	heta)$.$6	heta - 60 = 770 - 11	heta$.$17	heta = 830$.$	heta = \frac{830}{17} \approx 48.82^{\circ}C$.Rounding to the nearest integer,the final temperature is approximately $49^{\circ}C$.

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