Question

A refrigerator converts 100 g of water at 20°C to ice at −10°C in 35 minutes. Calculate the average rate of heat extraction in terms of watts.

Given: Specific heat capacity of ice = 2.1 J g⁻¹°C⁻¹

Specific heat capacity of water = 4.2 J g⁻¹°C⁻¹

Specific latent heat of fusion of ice = 336 J g⁻¹

Answer 1

Given: Mass of water, \(m=100\,\text{g}\)

Initial temperature = \(20^\circ\text{C}\)

Final temperature = \(-10^\circ\text{C}\)

Specific heat capacity of water, \(c_w=4.2\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)

Specific heat capacity of ice, \(c_i=2.1\,\text{J g}^{-1}\,^\circ\text{C}^{-1}\)

Specific latent heat of fusion of ice, \(L=336\,\text{J g}^{-1}\)

Time taken, \(t=35\times60=2100\,\text{s}\)

Heat extracted to cool water from 20°C to 0°C.

\[Q_1=mc_w\Delta T \]

\[=100\times4.2\times20 \]

\[=8400\,\text{J} \]

Heat extracted to convert water at 0°C into ice at 0°C.

\[ Q_2=mL \]

\[ =100\times336 \]

\[ =33600\,\text{J} \]

Heat extracted to cool ice from 0°C to −10°C.

\[ Q_3=mc_i\Delta T \]

\[ =100\times2.1\times10 \]

\[ =2100\,\text{J} \]

Hence, the total heat released.:

Q = Q₁ + Q₂ + Q₃

=  8400 + 33600 + 2100

= 44100 J

Therefore, the average rate of heat extraction.

\[ P=\frac{\text{Total heat extracted}}{\text{Time taken}} \]

= `44100/2100` 

= 21 W