One way to keep the contents of a garage from becoming too cold on a night when a severe subfreezing temperature is forecast is to put a tub of water in the garage. If the mass of the water is $\mathbf{125}\mathbf{}\mathbf{kg}\mathbf{}$and its initial temperature is$\mathbf{20}\mathbf{°}\mathbf{C}\mathbf{}$, (a) how much energy must the water transfer to its surroundings in order to freeze completely and (b) what is the lowest possible temperature of the water and its surroundings until that happens?

1. Mass of the water,$\mathrm{m}=125\text{\hspace{0.17em}\hspace{0.17em}kg}$
2. Initial temperature,${\mathrm{T}}_{\mathrm{i}}={20}^0\mathrm{C}$
3. Specific heat capacity of water,$\mathrm{c}=4190\text{\hspace{0.17em}J/kg.K}$
4. Specific latent heat of water,${\mathrm{L}}_{\mathrm{F}}=333\times {10}^3\text{J/kg}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. We can use the concept of specific heat of the water and change of state. The heat transformation involved in the phase change from liquid to solid is called heat of fusion.

Formulae:

The heat energy required by a body,$\mathrm{Q}=\mathrm{mc\Delta T}$ …(i)

Where, m = mass

c = specific heat capacity

$\mathrm{\Delta T}$= change in temperature

Q= required heat energy

The heat energy released or absorbed by the body,$\mathrm{Q}=\mathrm{Lm}$ …(ii)

Where, m = mass

L = specific latent heat

The energy transferred by the water to the surrounding in order to freeze completely:

The initial temperature of the water is $20°\mathrm{C}$. In order to freeze the water, energy is released by water in two steps. First, the released energy decreases the temperature from$20°C$to $0°C$, therefore, the energy transferred by the water is given as:

${\mathrm{Q}}_1=\mathrm{cm}({\mathrm{T}}_{\mathrm{f}}-{\mathrm{T}}_{\mathrm{i}})$

When the water freezes completely, then there is phase change and the energy transferred by the water is${\mathrm{Q}}_2={\mathrm{L}}_{\mathrm{F}}\mathrm{m}$

Hence, the total energy transferred to the surroundings is given as:

$\begin{array}{c}\mathrm{Q}={\mathrm{Q}}_1+{\mathrm{Q}}_2\\ =4190\text{J/kg.K}\times 125\text{kg}\times (20°\mathrm{C}-0°\mathrm{C})+333\times {10}^3\text{J/kg}\times 125\text{kg}\\ =5.2\times {10}^7\text{J}\end{array}$

Hence, the value of the required energy is$5.2\times {10}^7\text{J}$

The lowest temperature of the water before all the water freezes is given as:${\mathrm{T}}_{\mathrm{Low}}=0°\mathrm{C}$

Hence, the value of the lowest temperature is$0°\mathrm{C}$