A metal ball of mass 0.1 , kg is heated up to | vedclass

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(D) Let the final equilibrium temperature be $T$. According to the principle of calorimetry, heat lost by the metal ball = heat gained by the water an

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(D) Let the final equilibrium temperature be $T$. According to the principle of calorimetry, heat lost by the metal ball = heat gained by the water and the vessel. $m_{metal} c_{metal} (T_{initial, metal} - T) = (m_{water} c_{water} + C_{vessel}) (T - T_{initial, water})$ $0.1 	imes 400 	imes (500 - T) = (0.5 	imes 4200 + 800) 	imes (T - 30)$ $40(500 - T) = (2100 + 800) 	imes (T - 30)$ $20000 - 40T = 2900(T - 30)$ $20000 - 40T = 2900T - 87000$ $107000 = 2940T$ $T = \frac{107000}{2940} \approx 36.39 \, ^\circ C$ Change in temperature $\Delta T = 36.39 - 30 = 6.39 \, ^\circ C$. Percentage increment = $\frac{\Delta T}{T_{initial}} 	imes 100 = \frac{6.39}{30} 	imes 100 \approx 21.3 \%$. Rounding to the nearest given option, the answer is $20 \%$.

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