Question

calculate the amount of work that must be done to freeze one gram of water at zero degree celsius by using a refrigerator if the machine absorbs 80 Calories of heat from it the temperature of the surrounding is 27 degree Celsius​

Answer 1

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Q2=mL=1×80=80cal

T2=0∘C=273K

and

T1=27∘C=300K

Least amount of work will be needed for carnot's type of cycle.

Q2W=T2T1−T2

∴ W=Q2(T1−T2)T2

=80(300−273)273

=7.91cal

Q1=Q2+W

=(80+7.91)

=87.91cal