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

Asked by Pragya Singh | 1 year ago |  187

1 Answer

Solution :-

Mass of the copper block, m = 2.5 kg

Temperature of the block, ΔT= 500°C

Specific heat of copper, c= 0.39 J g–1 K–1

Latent Heat of fusion of water, L = 335 J g–1

Let m’ be the mass of the ice melted

Therefore, heat gained by ice = heat lost by copper

m’L = mcΔT

m’ =\( \dfrac{mc ΔT}{L}\)

\( \dfrac{(2500 \times 0.39 \times 500)}{335}\) = 1500 g = 1.5 kg

Answered by Pragya Singh | 1 year ago

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