The amount of heat energy required to convert 1 kg of ice at -10oC completely into water at 100oC is 777000 J. calculate the specific latent heat of ice.

Home » The amount of heat energy required to convert 1 kg of ice at -10oC completely into water at 100oC is 777000 J. calculate the specific latent heat of ice.

The amount of heat energy required to convert 1 kg of ice at -10^oC completely into water at 100^oC is 777000 J. calculate the specific latent heat of ice.

Calorimetry | Class 10 | ICSE | Physics | Selina

The amount of heat energy required to convert 1 kg of ice at -10^oC completely into water at 100^oC is 777000 J. calculate the specific latent heat of ice.

Specific heat capacity of ice = 2100 J kg^-1K^-1, Specific heat capacity of water is 4200 J kg^-1K^-1.

Solution:

According to the question, 1 kg of ice at -10^oC is completely converted into water at 100^oC. Using the expression for heat energy i.e., Q = m × c × (change in temperature), we can write the expression for heat energy gained by 1 kg of ice at – 10⁰ C in order to raise its temperature to 0⁰ C is

= 1 × 2100 × 10 = 2100 J

Similarly, heat energy gained by 1 kg of ice at 0⁰ C in order to convert into water at 0⁰ C = L

Then, the heat energy gained when the temperature of 1 kg of water at 0⁰ C rises to 100⁰ C is

= 1 × 4200 × 100 = 420000 J

Therefore, the total amount of heat energy gained

= 21000 + 420000 + L = 441000 + L

It is given that the total amount of heat gained is 777000 J

So, 441000 + L = 777000

Or, we can write => L = 777000 – 441000

Thus, L = 336000 J kg^-1

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