Question

0.5 kg of lemon squash at ${30}^{0}C$ is placed in a refrigerator which can remove heat at an average rate of 30 $J{s}^{-1}$. How long will it take to cool the lemon squash to $5^{0}C$? (Specific heat capacity of squash = 4200 J $K{g}^{-1}{K}^{-1})$.

• A. 20 sec
• B. 30.5 sec
• C. 25 min
• D. 29.2 min

Answer 1

The correct option is D

29.2 min

m = 0.5 kg

Fall in temperature, $T=(30–5)={25}^{0}C$

Specific Heat Capacity of squash, $C=4200JK{g}^{-1}{}^0{C}^{-1}$​

Heat removed per second = 30 J

Let t second is time,

t = $\frac{mcT}{30}$ = 0.5 $\times$ 4200 $\times$ $\frac{25}{30}$ = 1750 seconds

t = $\frac{1750}{60}$ = 29.2 min