Question

A refrigerator works between 4 C and 30 C. It is required to remove 600 of calories heat every second in order to keep temperature of the refrigerated space constant. The power required is (take 1 cal.=4.2 joule)

a.2.365 W
b.23.65 W
c.236.5 W
d.2365 W
d.

Answer 1

Solution is attached. Please see!

Answer 2

The required power is 236.5 W

Given:

Temperature of the source, $T_{1}$= 30 C = 30 + 273 = 303 K

Temperature of sink, $T_{2}$= 4 C = 4 + 273 = 277 K

To find:

Power required for the given condition

Solution:

We Know,

$\frac{Q_{1}}{Q_{2}}=\frac{T_{1}}{T_{2}}$

$\frac{Q_{2}+W}{Q_{2}}=\frac{T_{1}}{T_{2}}$

$W= Q_{1} - Q_{2}$

Where, Q2 is the heat drawn out from the sink at T2

W is the work done on the working substance,

Q1 is the heat rejected to source at room temperature T1.

 That is,

$W T_{2}+T_{2} Q_{2}=T_{1} Q_{2}$

$W T_{2}=T_{1} Q_{2}-T_{2} Q_{2}$

$W T_{2}=\left(T_{1}-T_{2}\right) Q_{2}$

$W=Q_{2} \frac{\left(T_{1}-1\right)}{T_{2}}$

$W=600 \times 4.2\left(\frac{303-1}{277}\right)$

$W=600 \times 4.2\left(\frac{26}{277}\right)$

W = 236.5 J

Power $=\frac{\text {Work done}}{\text {time}}$

$=\frac{236.5}{1}$

=236.5 J