Question #ae1f7

1 Answer
Bharat K.
Apr 28, 2015

Required parameters to solve the question:

  1. Specific heat of water, S = 4.1 J K^(-1) g^(-1)
  2. Latent heat of fusion of water, L = 336 J g^(-1)

Now, the amount of heat which should be released in order to convert water at 26^0 C to ice at -15^0 C is given by

Q = mS(T_h -T_c) + mL

where m is the mass of water.

Substituting all the values in above equation, we get

Q = 500( 4.1*(26+15) + 336)) = 252.05 KJ

Now the work required by the refrigerator to convert water at 26^0 C to ice at -15^0 C must be equal to the heat which needs to be released to perform the same conversion.

Therefore, W = 252.05 KJ

Now, P_(f ridg e) = 230W

and we know that, P = W/t

therefore, t = W/P = (252.05 * 1000)/230 = 1095.8 s = 18.26 min

Related questions
Impact of this question
993 views around the world
You can reuse this answer
Creative Commons License