A cube of iron (density = 8000 kg m–3, specific heat capacity = 470 J kg–1 K–1) is heated to a high temperature and is placed on a large block of ice at 0°C. The cube melts the ice below it, displaces the water and sinks. In the final equilibrium position, its upper surface just goes inside the ice. Calculate the initial temperature of the cube. Neglect any loss of heat outside the ice and the cube. The density of ice = 900 kg m–3 and the latent heat of fusion of ice = 3.36 × 105 J kg–1.

Given:


iron (density = 8000 kg m–3, specific heat capacity = 470 J kg–1 K–1)


density of ice = 900 kg m–3


latent heat of fusion of ice = 3.36 × 105 J kg–1.


Formula used:


ΔQ=MCΔT


ΔQ=heat exchange


ΔT=tempeature change


M=mass


C=heat capacity


Heat lost by iron= Heat gained by ice


8000× Volume× 470× T=900× Volume × 3.36× 105


T=80.42° C


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