Power input of the refrigerator is #1 kW# means per second is supplies #1*10^3J# of energy
So,in #1 hr# it will supply #10^3 *3600=36*10^5 J=8.56*10^5# calories of heat energy.
Now,let the refrigerator will take the temperature of water to #T_2# and the water is at #T_1# temperature.
Given, #beta =7,T_1=0+273=273K#
So,using the equation, #beta =(T_2)/(T_1 -T_2)#
We,get, #T_2=238.875K=-34.125^@C#
Now,heat energy required for conversion of #m# # gram# of water at #0^@C# to same amount of ice at #0^@C# i.e the latent heat required is #80m# (as,latent heat of water is #80#calories per gram)
So,to take this #m# #gram# of ice at #0^@C# to #-34.125^@C# heat energy required will be #m*0.5*(0-(-34.125))# calories (using the formula, heat required #H=ms d theta#,where, #s# is the specific heat,for ice it is #0.5# C.G.S units and #d theta# is the change in temperature)
So,this entire amount of heat will be equals to #8.56*10^5# calories
So, #80m+m*0.5*34.125=8.56*10^5#
we get,from the above equation, #m=8819.06# #gram#
Now,density of water is #1# #(gram)/(cm^3)#
So,volume of this much water will be #8819.06/1=8819.06 cm^3# (as, #v=m/d#)
Now, #1 cm^3=0.001 L#
so, #8819.06 cm^3=8.81906=8.82L#
