The inside and outside temperatures of a refrigerator are 273 K and 303 K respectively. Assuming that refrigerator cycle is reversible, for every joule of work done the heat delivered to the surrounding will be

Question

The inside and outside temperatures of a refrigerator are 273 K and 303 K respectively. Assuming that refrigerator cycle is reversible, for every joule of work done the heat delivered to the surrounding will be (A) 10 J (B) 20 J (C) 30 J (D) 50 J

Tardigrade Admin · Accepted Answer

β=(Q2/W)=(TL/TH-TL) Q2 =(273 × 1/303-273)=(273/30)=9 J Heat delivered to the surrounding Q1 =Q2+W =9 + 1 = 10 J

Q. The inside and outside temperatures of a refrigerator are $273 K$ and $303 K$ respectively. Assuming that refrigerator cycle is reversible, for every joule of work done the heat delivered to the surrounding will be

10 J

20 J

30 J

50 J

$β = \frac{Q _{2}}{W} = \frac{T _{L}}{T _{H} - T _{L}}$
$Q_{2} = \frac{273 \times 1}{303 - 273} = \frac{273}{30} = 9 J$
Heat delivered to the surrounding
$Q_{1} = Q_{2} + W$
$= 9 + 1 = 10 J$

Q. The inside and outside temperatures of a refrigerator are 273K and 303K respectively. Assuming that refrigerator cycle is reversible, for every joule of work done the heat delivered to the surrounding will be