10 grams of ice at 0^oC absorbs 5460 J of heat energy to melt and change the water to 50^oC. Calculate the specific latent heat of fusion of ice.The specific heat capacity of water is 4200 J / KG K.

Given:

Mass of ice, m = 10 g = 0.01 kg [converted gram to kilogram] 

Amount of heat energy absorbed, Q = 5460 J

Specific heat capacity of water = 4200Jkg^-1K^-1

Specific latent heat of fusion of ice, L = ? 

Solution: 

Amount of heat energy required by 10g (0.01kg) of water at 0^oC to raise its temperature by 50^oC= 0.01 x 4200 x 50 = 2100J.

Let Specific latent heat of fusion of ice = L Jg^-1.

We know that the expression for the heat energy Q can be given as-

$Q=mL+mc\triangle T$

Substituting the values in the expression we get-

5460 J =10 x L + 2100 J

L = 336Jg^-1

Hence,  the specific latent heat of fusion of ice will be 336Jg^-1.

Tutorialspoint

e

Updated on: 10-Oct-2022

1K+ Views

Kickstart Your Career

Get certified by completing the course

Get Started