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(C) The heat lost by the copper block as it cools from $500^{\circ} C$ to $0^{\circ} C$ is given by the formula $\Delta Q = mc\Delta T$.Mass of copper

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$2.9$

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(C) The heat lost by the copper block as it cools from $500^{\circ} C$ to $0^{\circ} C$ is given by the formula $\Delta Q = mc\Delta T$.Mass of copper $m = 5.0 \, kg = 5000 \, g$.Specific heat of copper $c = 0.39 \, J g^{-1} {}^{\circ} C^{-1}$.Temperature change $\Delta T = 500^{\circ} C - 0^{\circ} C = 500^{\circ} C$.Heat released $\Delta Q_1 = 5000 	imes 0.39 	imes 500 = 975,000 \, J$.Let $m_{ice}$ be the mass of ice that melts. The heat absorbed by the ice to melt at $0^{\circ} C$ is $\Delta Q_2 = m_{ice} 	imes L_f$,where $L_f = 335 \, J g^{-1}$.Equating the heat lost and heat gained: $\Delta Q_1 = \Delta Q_2$.$975,000 = m_{ice} 	imes 335$.$m_{ice} = \frac{975,000}{335} \approx 2910.45 \, g$.Converting to $kg$,$m_{ice} \approx 2.91 \, kg$.

$A$ copper block of mass $5.0 \, kg$ is heated to a temperature of $500^{\circ} C$ and is placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt? [Specific heat of copper: $0.39 \, J g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of water: $335 \, J g^{-1}$]

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