An ideal refrigerator has a freezer at a temperature of -13° C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is

Question

An ideal refrigerator has a freezer at a temperature of -13° C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is (A) 320° C (B) 39° C (C) 325 K (D) 325° C

Tardigrade Admin · Accepted Answer

Given: T H =-13° C β=5 ∴ T H =273-13=260 K Coefficient of performance β=( T H / T H - T L ) ∴ 5=(260/260- T L ) ⇒ T L =312 K Thus temperature of air TL=312-273=39° C

Q. An ideal refrigerator has a freezer at a temperature of $- 1 3^{\circ} C$ . The coefficient of performance of the engine is $5$ . The temperature of the air (to which heat is rejected) is

$32 0^{\circ} C$

$3 9^{\circ} C$

$325 K$

$32 5^{\circ} C$

Given: $T_{H} = - 1 3^{\circ} C β = 5$
$∴ T_{H} = 273 - 13 = 260 K$
Coefficient of performance
$β = \frac{T _{H}}{T _{H} - T _{L}}$
$∴ 5 = \frac{260}{260 - T _{L}}$
$\Rightarrow T_{L} = 312 K$
Thus temperature of air
$T_{L} = 312 - 273 = 3 9^{\circ} C$

Q. An ideal refrigerator has a freezer at a temperature of −13∘C. The coefficient of performance of the engine is 5. The temperature of the air (to which heat is rejected) is

320∘C

39∘C

325K

325∘C