Question

what amount of ice will remain when 52g ice is added to 100g water at 40 C. specefic heat of water is 1cal/g and latent of fusion of ice is 80 cal/g.

Answer 1

2 grams of ice will remain.

Thermal equilibrium of the system of the system will be achieved at 0°C.

At thermal equilibrium -

The Heat lost by water = Heat absorbed by ice to melt

=> msΔT=m'L

m - mass of the water = 100 g

m' - mass of the ice melted

∆T - The change in the temperature of water = 40°C

s - specific heat of water = 1 Cal/g

L - latent heat of fusion = 80 Cal/g

=> 100×1×40=m×80

=> m=50g

Remaining ice = 52-50

= 2g