Question

Suppose you had some 0o C water in a freezer with a COP of 6.2 that is receiving 1.25 kW of power from the wall. How much mass of water could this thing freeze in an hour?

Answer 1

The COP of a freezer is defined as
   $\beta =\frac{Q_c}{W}$

where,
c
is the heat removed from the inside of the freezer. And W is the electrical energy supplied to the system from the wall outlet. Now if m amount of water is frozen and L is the latent heat, then, the heat removed is
   
Tu=0 c

So, we have
   
m. 8 = LL W

  
BW m L

So, the rate at which the water is frozen is
  
В dW dm L dt dt

where,
   
MP dt 1P

  is the power of the wall outlet.
So, we get
  
dm BP dt

Now using the standard and given values,
   
86.2, P 1.25 x 103 watts, L 334 x 103 J/kg

we get the rate of freezing
  
6.2 x 1.25 x 103 kg/s dm dt 334 x 103

  
dm 0.0232 kg/s dt

So, in the kg/hr unit, we get
  
dm 0.0232 x 3600 kg/hr dt

  $\Rightarrow \frac{\mathrm{d} m}{\mathrm{d} t} =83.53~kg/hr$
So, 83.53 kg of water will freeze in an hour.