A refrigerator converts 100 g of water at 20oC to ice at -10oC in 73.5 min. calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g-1 K-1, specific latent heat of ice is 336 J g-1 and the specific heat capacity of ice is 2.1 J g-1 K-1.
A refrigerator converts 100 g of water at 20oC to ice at -10oC in 73.5 min. calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g-1 K-1, specific latent heat of ice is 336 J g-1 and the specific heat capacity of ice is 2.1 J g-1 K-1.

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