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

Question

A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is: (Take 1 cal = 4.2 Joules) (A) 23.65 W (B) 236.5 W (C) 2365 W (D) 2.365 W

Tardigrade Admin · Accepted Answer

β = (Q2/W) = (T2/T1 - T2) (Where Q2 is heat removed) ⇒ (600 × 4.2/W) = (277/303 - 277) ⇒ W = 236.5 joule ⇒ Power = (W/t) = (236.5 joule/1 sec) = 236.5 watt.

Q. A refrigerator works between $4^{\circ} C$ and $3 0^{\circ} C$ . It is required to remove $600 c a l or i es$ of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : $(T ak e 1 c a l = 4.2 J o u l es)$

23.65 W

236.5 W

2365 W

2.365 W

$β = \frac{Q _{2}}{W} = \frac{T _{2}}{T _{1} - T _{2}}$ (Where $Q_{2}$ is heat removed)
$\Rightarrow \frac{600 \times 4.2}{W} = \frac{277}{303 - 277}$
$\Rightarrow W = 236.5$ joule
$\Rightarrow P o w er = \frac{W}{t} = \frac{236.5 j o u l e}{1 sec} = 236.5 w a tt .$

Q. A refrigerator works between 4∘C and 30∘C. It is required to remove 600calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is : (Take1cal=4.2Joules)