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Home » 200 g of ice at 0 °C converts into water at 0 °C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0 °C will change to 20 °C? Take specific latent heat of ice = 336 J g-1.
200 g of ice at 0 °C converts into water at 0 °C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0 °C will change to 20 °C? Take specific latent heat of ice = 336 J g-1.
Calorimetry | Class 10 | ICSE | Physics | Selina
200 g of ice at 0 °C converts into water at 0 °C in 1 minute when heat is supplied to it at a constant rate. In how much time, 200 g of water at 0 °C will change to 20 °C? Take specific latent heat of ice = 336 J g-1.

Solution:

According to the question, mass of ice is mice= 200 g

Time (t1) for ice to melt is 1 min, or 60 s

Mass of water is mw = 200 g

Temperature change of water is ΔT = 200 C

Heat exchange occurs at a steady rate. As a result, the power required to convert ice to water is the same as the power required to raise the water’s temperature.

Therefore, Pice = Pwater

We know that the expression for Power is ==> P =E/t

So, we obtain => Eice / t1 = Ewater / t2

Expanding we get => miceL / t1 = mwcw ΔT / t2

Solving for t2 , we get => t2 = (mwcw ΔT × t1) / miceL

t2 = (200 × 4.2 × 20 × 60) / 200 × 336

Therefore, t2 = 15 s

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