Question

A copper block of mass 2.5 kg is heated in a furnace to a temp of 500° C and then placed on a large ice block. What is the maximum amount of ice that can melt?​

Answer 1

By law of mixtures,

Heat given by copper block = Heat taken by ice

Mc◇T=m ice L

=> m ice = Mc◇T/L

Given M=2.5kg,◇T=(500-0)=500°C

Specific heat of copper,c=0.39Jkg^-1 K^-1

Heat of fusion of water , L= 335Jg-1=335×10^3 Jkg^-1

m ice =

$\frac{2.5 \times 0.39 \times 10 ^{3} \times 500}{335 \times 10 ^{3} } = 1.455 \: kg$

=1.5 kg

Answer 2

Answer:

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