Heat gained by water of 0.4kg to change its temperature=q1
q1=mcΔT
Where,
m=mass
C= Specific capacity
ΔT= Temperature difference
q1=0.4×103×40×1
=16000 calories (specific heat of water= 1cal/gm°C)
Heat gained by0.4kg of ice to change its temperature =q2
q2=mcΔT (Specific capacity of ice=12cal/gm°C)
=0.1×12×15×1000
=0.75calories
=750calories
Heat gained by .01kg of ice to convert into water=mL
q3=mL (L=Latent heat of fusion)
=0.1×80×1000
= 8000calories
Heat gained by 0.14kg water to change temperature from 0°C to 40°C=mcΔT=q4
q4=0.110×1000×1×40
=4000calories
Total heat gained =q1+q2+q3+q4
=16000+750+8000+4000
=28750calories
Heat lost by steam to change state to liquid=q5
q5=mL (Latent heat of vaporization)
q5=m×540calories
Heat lost by water to change temperature from 100°C to 40°C= q6=mcΔT
=m×1×60
=60m calories
At equilibrium heat gained = heat lost
28750=600m
m=28750600gm
m=48.01gm