10 gm of ice at -20°C is added to 10gm of water at 50°C. Specific heat of water = 1 cal / gm /°C, specific heat of ice = 0.5 cal / gm /° C. Latent heat of ice = 80 cal/gm. Then, resulting temperature is

Solution:

Let resulting temperature = 0°C
$\therefore$ Q₁ = heat given by water if it was to cool upto 0°C
= 10.1(50 - 0) = 500 cal and Q₂ = heat required by ice to convert totally into water at 0°C .
= heat required to raise the temperature of ice from
- 20°C to 0°C + heat required to melt 10 gm of ice at
0°C into water at 0°C .
= 100 + 800 = 900 cal
As Q₁ < Q₂, hence whole of the ice cannot melt.
Initially, 100cal of heat will be used up in raising the temperature of ice to 0°C and the rest 400 cal will be available for melting of ice. If it melts m gm of ice, then
m' 80 = 400, i.e., m' = 5 gm
Hence, only 5 gm of ice will melt and the remaining 5 gm of ice will remain in the mixture as ice at 0°C. The amount of water in the mixture.
10 gm + 5 gm = 15 gm
Final temperature of mixture = 0°C.