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(A) Mass of the copper block,$m = 2.5\; kg = 2500\; g$.Change in temperature of the copper block,$\Delta 	heta = 500\; ^{\circ}C$.Specific heat of co

Vedclass · Accepted Answer

$1.45$

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(A) Mass of the copper block,$m = 2.5\; kg = 2500\; g$.Change in temperature of the copper block,$\Delta 	heta = 500\; ^{\circ}C$.Specific heat of copper,$C = 0.39\; J\; g^{-1}\; ^{\circ}C^{-1}$.Heat of fusion of water,$L = 335\; J\; g^{-1}$.The maximum heat the copper block can lose,$Q = m C \Delta 	heta$.$Q = 2500\; g 	imes 0.39\; J\; g^{-1}\; ^{\circ}C^{-1} 	imes 500\; ^{\circ}C = 487500\; J$.Let $m_1$ be the mass of ice that melts. The heat gained by the ice to melt is $Q = m_1 L$.$m_1 = \frac{Q}{L} = \frac{487500\; J}{335\; J\; g^{-1}} \approx 1455.22\; g$.Converting to kilograms,$m_1 \approx 1.45\; kg$.

$A$ copper block of mass $2.5\; kg$ is heated in a furnace to a temperature of $500\; ^{\circ}C$ and then placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt? (Specific heat of copper $= 0.39\; J\; g^{-1}\; K^{-1}$; heat of fusion of water $= 335\; J\; g^{-1}$)

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