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10 g of ice at 0oC absorbs 5460 J of heat energy to melt and change to water at 50oC. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.
10 g of ice at 0oC absorbs 5460 J of heat energy to melt and change to water at 50oC. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.
Calorimetry | Class 10 | ICSE | Physics | Selina
10 g of ice at 0oC absorbs 5460 J of heat energy to melt and change to water at 50oC. Calculate the specific latent heat of fusion of ice. Specific heat capacity of water is 4200 J kg-1 K-1.

Solution:

According to the question, Mass of ice = 10 g = 0.01 kg

Amount of absorbed heat energy is Q = 5460 J

We have to find the specific latent heat of fusion of ice.

We are given that the specific heat capacity of water = 4200 J kg-1 K-1

Using the expression of heat energy => Q = mc(change in temperature), we can write the expression for amount of heat energy required by 10 g (0.01 kg) of water at 00 C to raise its temperature by 500 C.

Therefore, Heat Energy = 0.01 × 4200 × 50 = 2100 J

Let the specific latent heat of fusion of ice be L Jg-1

Then using the other expression of heat energy, i.e., Q = mL + mc△t

Putting values in above equation, we get

=> 5460 J = 10 × L + 2100 J

Or, L = 3360 / 10

Therefore, L = 336 J g-1

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