Question

A refrigerator converts $100g$ of water at ${20}^{o}C$ to ice at $-{10}^{o}C$ in $35$ minutes. Calculate the average rate of heat extraction in terms of watts.
Given:
Specific heat capacity of ice $=2.1J{g}^{-1}{}^o{C}^{-1}$
Specific heat capacity of water $=4.2J{g}^{-1}{}^o{C}^{-1}$
Specific Latent heat of fusion of ice $336J$ $g^{-1}$

Answer 1

Mass of water $=100g=0.1kg$

Temperature of water $={20}^{o}C$

Amount of heat extracted to convert water from ${20}^{o}C$ to $0^{o}C$.

$Q_1=m{C}_{water}\Delta t=0.1\times 4200\times (20-0)$ $=8400J$

Amount of heat extracted to convert water at $0^{o}C$ to $0^{o}C$ ice

$Q_2=m{L}_{ice}=0.1\times 336\times 1000$ $=33600J$

Amount of heat extracted to convert ice at $0^{o}C$ to ice at $-{10}^{o}C$

$Q_3=m{C}_{ice}\Delta t=0.1\times 2.1\times {10}^3\times (0-(-10\left)\right)$ $=2100J$

Total Heat $\left(Q\right)=Q_1+Q_2+Q_3$

$=8400+33600+2100=44100J$

$\therefore$ $Pt=Q$

$P=\frac{q}{t}$

Power $=\frac{44100}{35\times 60}=\frac{44100}{2100}=21W$