Question

100g ice at 0°C is informally heated to convert it into water at 50°C. Calculate the amount of heat taken by ice and water in joules.

Given: Latent heat of fusion of ice = 80 Calorie / g Specific thermal capacity of water = 1 Calorie g-10 c-1 Calorie = 4.2 J/Cal.

Answer 1

when we heated ice and they melt heat =mLf
where Lf is latent heat of fusion
100gx80cal/g=8000 cal
now formed water heat at 50 degree C
heat=msdT
where s is specific heat of water
dT is change in temperature
heat=100x1x (50-0)=5000cal
so total heat=8000+5000=13000 cal=13000x4.2 j