Answer :

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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