A refrigerator converts 5 kg of water at$40\text{°C}$  into ice at$0\text{°C}$  in 20 minutes. Then its power is

m = 5 kg of water ; t = 40 °C

convert ice at 0°C

Energy absorbed

$\text{Q}=\text{ms}\Delta \text{t}+\text{mL}$

$=\text{5}\times \text{4200}\times \text{40}+\text{5}\times \text{80}\times \text{4200}=\text{4200}\times \text{5}\times \text{40}\times \left(1+2\right)$

$=4200\times 5\times 40\times 3\text{J}$

$\text{t}=\text{20min}=\text{20}\times \text{60s}$

$\text{p}=\frac{4200\times 5\times 40\times 3}{20\times 60}=2100\text{watts}$