Question

when 0.1 kg of ice at 0°c is mixed with 0.32 kg of water at 35°c in the mixture is 78° c calculate the heat of fusion of ice.​

Answer 1

Answer:

for ice :

heat required for melting = mLf

where m is mass of ice and Lf is Latent heat of fusion.

and heat required to change to the temperature of water from 0°C to 50°C

= mS(50 - 0) = 50mS

where S is specific heat of water .

we know,

Heat Loss = Heat gain,

5460 Joule = mLf + 50mS

5460 = 10× 10^-3 Lf + 50 × 10× 10^-3 × 4200

5460 - 2100 = 10^-2 Lf

Lf = 3360× 10² = 3.36 × 10^5 j/kg