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.
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

Step 1:Given that
mass of water $m_w=100g$
Specific heat capacity of water $s_w=4.2J{g}^{-1}{}^o{C}^{-1}$
Temperature of water $T={20}^{o}C$
Specific Latent heat of fusion of ice $L=336J{g}^{-1}$
Specific heat capacity of ice $s_w=2.1J{g}^{-1}{}^o{C}^{-1}$
Temperature of ice $T\text{'}=-{10}^{o}C$

First, the water at ${20}^{o}C$is taken to the $0^{o}C$temperature. After that latent heat is required to convert water to ice and then it is reached at $-{10}^{o}C$.

Step 2:Calculate heat needed
$Q=m_w{s}_w(T-0)+m_{w}L+m_w{s}_i(0-T\text{'})$

or,$Q=\text{100}\times \text{4.2}\times 20+100\times 336+100\times 2.1\times 10$
or, $Q=44.1 kJ$

Step 3:Calculate the average rate of heat extraction

$Q=44.1 kJ$ heat is extracted in $35$minutes, that is, $35\times 60=2100s$
So, heat extraction rate per second $=\frac{44100}{2100}=21J/s=21 W$
In the given situation, heat extraction rate in watts is $21$.