A copper ball of mass 100gm is at a temperature T. It is dropped in a copper calorimeter of mass 100gm, filled with 170gm of water at room temperature. Subsequently, the temperature of the system is found to be 75oC. T is given by :(Given : room temperature = 30oC, specific heat of copper = 0.1cal/gmoC)

The amount of heat lost by copper ball = mball * CCu * ΔTball = (100gm) * (0.1cal/gm°C) * (T - 75)°C
The amount of heat gained by calorimeter = mcalorimeter * CCu * ΔTCalorimeter = (100gm) * (0.1cal/gm°C) * (75 - 30)°C
Amount of heat gained by water = mw * Cw * ΔTw = (170gm) * (1cal/gm°C) * (75 - 30)°C
Hence, from conservation of energy,
100 * 0.1 * (T - 75) = 100 * 0.1 * (75 - 30) + 170 * 1 * (75 - 30)
=> 10(T - 75) = 450 + 7650
=> 10(T - 75) = 8100
=> T - 75 = 810
=> T = 885°C
Hence, the correct answer is option B.