Question

A refrigerator converts 100 g of water at ${20}^{o}C$ to ice at $-{10}^{o}C$ in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J $g^{-1}{K}^{-1}$, specific latent heat of ice is 336 J $g^{-}1$ and the specific heat capacity of ice is 2.1 J $g^{-1}{K}^{-1}$.

Answer 1

Heat extracted from water at ${20}^{o}C$ to convert it to ice at $-{10}^{o}C$ = $m{S}_w\Delta T_w+m{L}_f+m{S}_i\Delta T_i$

$=100\times 4.2\times 20+100\times 336+100\times 2.1\times 10$

$=8400+33600+2100=44100J$

Time taken = 73.5 min = 73.5 $\times$ 60 s = 4410 s

So, rate of heat extraction = $\frac{44100}{4410}=10J/s=10W$