Solution:
According to the question, mass of water is 100g at 20 degrees C is converted to ice at -10 degrees C in 73.5 mins. We will first calculate the heat energy by using the following expression –
Q = m × c × × (change in temperature)
We have, specific heat capacity of water = 4.2 J g-1 K-1,
specific latent heat of ice = 336 J g-1 and
the specific heat capacity of ice = 2.1 J g-1 K-1
So, using above expression the amount of heat released when 100 g of water cools from 200 to 00 C is
= 100 × 20 × 4.2 = 8400 J
Similarly, the amount of heat released when 100 g of water converts into ice at 00 C
= 100 × 336 = 33600 J
Again, the amount of heat released when 100 g of ice cools from 00 C to -100 C
= 100 × 10 × 2.1 = 2100 J
Therefore, the total amount of heat energy released = 8400 + 33600 + 2100
Total Heat Energy= 44100 J
We know that the time is 73.5 min, or 4410 s
Using the expression for average rate of heat extraction (power), i.e.,
P = E / t
Putting values, we get P = 44100 / 4410
Thus, P = 10 W