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Question

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.Specific heat of copper =0.39Jg^-1K^-1,heat of fusion of water =335Jg^-1 Maximum amount of ice that can be melted

A)1 kg.     B)1.5 kg.    C) 2 kg.      D)2.5 kg


Solution

We know 
heat loss =heat gain 
maximum heat loss by copper block =mSdT
=2500g x 0.39jg-1k-1 x (500-0) k
=487,500 j 

now 
487500 j=M L
where M is the mass of ice melted and L is latent heat of fusion .
487500 j =M x 335jg-1
M=1455.22 g =1.45522 kg


So option is 1.5kg

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