Question

10g of ice at 0°C absorbs 5460J of heat energy to melt and change to water at 50°C. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200J per kg per Kelvin.

Answer 1

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