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#

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