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