The water equivalent of a calorimeter is 10 | vedclass

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(C) Let the initial mass of ice be $m \ g$.According to the principle of calorimetry, heat lost by the system (calorimeter + water) = heat gained by t

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$20$

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(C) Let the initial mass of ice be $m \ g$.According to the principle of calorimetry, heat lost by the system (calorimeter + water) = heat gained by the ice.The system cools down to $0^{\circ} C$ because only half of the ice melts, implying the final equilibrium temperature is $0^{\circ} C$.Heat lost by water and calorimeter: $Q_{lost} = (m_{water} + m_{cal}) \cdot s_{water} \cdot \Delta T = (50 + 10) \cdot 1.0 \cdot (15 - 0) = 60 \cdot 15 = 900 \ cal$.Heat gained by ice: The ice warms from $-10^{\circ} C$ to $0^{\circ} C$ and then half of it melts.$Q_{gained} = m \cdot s_{ice} \cdot \Delta T_{ice} + (m/2) \cdot L_f = m \cdot 0.5 \cdot (0 - (-10)) + (m/2) \cdot 80$.$Q_{gained} = m \cdot 0.5 \cdot 10 + 40m = 5m + 40m = 45m$.Equating heat lost and gained: $900 = 45m$.$m = 900 / 45 = 20 \ g$.

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