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(B) According to the principle of calorimetry,heat lost by the copper ball is equal to the heat gained by the copper calorimeter and the water.Let $m_

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$885$

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(B) According to the principle of calorimetry,heat lost by the copper ball is equal to the heat gained by the copper calorimeter and the water.Let $m_b = 100 \ gm$ be the mass of the copper ball,$m_c = 100 \ gm$ be the mass of the calorimeter,and $m_w = 170 \ gm$ be the mass of water.Specific heat of copper $s_{Cu} = 0.1 \ cal/gm ^\circ C$ and specific heat of water $s_w = 1 \ cal/gm ^\circ C$.Initial temperature of the ball is $T$,and the final equilibrium temperature $T_f = 75 ^\circ C$. Initial temperature of the calorimeter and water $T_0 = 30 ^\circ C$.Heat lost by the ball $= m_b s_{Cu} (T - T_f) = 100 	imes 0.1 	imes (T - 75) = 10(T - 75)$.Heat gained by the calorimeter $= m_c s_{Cu} (T_f - T_0) = 100 	imes 0.1 	imes (75 - 30) = 10 	imes 45 = 450 \ cal$.Heat gained by the water $= m_w s_w (T_f - T_0) = 170 	imes 1 	imes (75 - 30) = 170 	imes 45 = 7650 \ cal$.Equating the heat lost and gained:$10(T - 75) = 450 + 7650$$10T - 750 = 8100$$10T = 8850$$T = 885 ^\circ C$.

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