(a) Two 50 gice cubes are dropped into 200 gof water in a thermally insulated container. If the water is initially at${\mathbf{25}}^{\mathbf{o}}\mathbf{C}$, and the ice comes directly from a freezer at$\mathbf{-}{\mathbf{15}}^{\mathbf{o}}\mathbf{C}$, what is the final temperature at thermal equilibrium? (b) What is the final temperature if only one ice cube is used?

1. Mass of the water,$\left({\mathrm{M}}_{\mathrm{W}}\right)=200\mathrm{g}$
2. Specific heat of water,$\left({\mathrm{C}}_{\mathrm{W}}\right)=4190\frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}$
3. Mass of ice,$\left({\mathrm{M}}_{\mathrm{I}}\right)=0.100\mathrm{kg}$
4. Specific heat of ice,$\left({\mathrm{C}}_{\mathrm{I}}\right)=2220\frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}$
5. Initial temperature of water,$\left({\mathrm{T}}_{\mathrm{Wi}}\right)={25}^{\mathrm{o}}\mathrm{C}$
6. Initial temperature of ice,$\left({\mathrm{T}}_{\mathrm{Ii}}\right)=-{15}^{\mathrm{o}}\mathrm{C}$

Specific heat is the quantity of heat that is required to raise the temperature of one gram of a substance by one Celsius degree. Here, the concept of specific heat is used to calculate the heat required by the hot water to be released that will be absorbed by the cold water in the further process. Considering three possible cases:

That is given as follows:

• No ice is melting and the water-ice system reaches thermal equilibrium at the temperature that is at or below the melting point of ice.
• The system reaches thermal equilibrium at the melting point of ice, with a decrease in mass of the ice with some melting.
• All of the ice melts and the system reaches thermal equilibrium at a temperature at or above the melting point of ice.

Thus, at equilibrium, we write that heat emitted by a hot object = heat absorbed by a cold object. Therefore, using the equation for the heat in terms of mass, specific heat, and temperature difference, we can find the final equilibrium temperature.

Formula:

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

Where, m = mass

c = specific heat capacity

$∆\mathrm{T}$= change in temperature

Q = required heat energy

First suppose that no ice melts. The temperature of the water decreases from$\left({\mathrm{T}}_{\mathrm{Wi}}\right)={25}^{\mathrm{o}}\mathrm{C}$ to some final temperature$\left({\mathrm{T}}_{\mathrm{f}}\right)$and the temperature of the ice increases from$\left({\mathrm{T}}_{\mathrm{Ii}}\right)=-{15}^{\mathrm{o}}\mathrm{C}$ to final temperature $\left({\mathrm{T}}_{\mathrm{f}}\right)$.

At equilibrium condition, we write the formula for heat using equation (i) as:

${\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}\left({\mathrm{T}}_{\mathrm{Wi}}-{\mathrm{T}}_{\mathrm{f}}\right)={\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}\left({\mathrm{T}}_{\mathrm{Ii}}-{\mathrm{T}}_{\mathrm{f}}\right)$

Then, the thermal equilibrium temperature is given as:

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{f}}& =& \frac{{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}{\mathrm{T}}_{\mathrm{Wi}}+{\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}{\mathrm{T}}_{\mathrm{Ii}}}{{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}+{\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}}\\ & =& \frac{4190{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.200\mathrm{kg}\times {25}^{\mathrm{o}}\mathrm{C}+2220{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.100\mathrm{kg}\times \left(-{15}^{\mathrm{o}}\mathrm{C}\right)}{4190{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.200\mathrm{kg}+2220{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.100\mathrm{kg}}\\ & =& \frac{20950-3330}{838+222}{}^{\mathrm{o}}\mathrm{C}\\ & =& \frac{17620}{1060}{}^{\mathrm{o}}\mathrm{C}\\ & =& 16.62{}^{\mathrm{o}}\mathrm{C}\end{array}$

Now, the energy required to warm all the ice is equal to the energy required to melt m mass of ice, so we can write

${\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}{\mathrm{T}}_{\mathrm{Wi}}=-{\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}{\mathrm{T}}_{\mathrm{Ii}}+{\mathrm{mL}}_{\mathrm{f}}$ …(ii)

Where, L_f is the heat of fusion of water. The first term is the energy required to warm all the ice from its initial temperature to 0^oC, and the second term is the energy required to melt mass m of ice. Therefore, the mass m of the ice using equation (I) is given as:

localid="1662393452711" $\begin{array}{rcl}\mathrm{m}& =& \frac{4190{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.200\mathrm{kg}\times {25}^{\mathrm{o}}\mathrm{C}+2220{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.100\mathrm{kg}\times \left(-{15}^{\mathrm{o}}\mathrm{C}\right)}{333\times {10}^3\mathrm{J}/\mathrm{kg}}\\ & =& \frac{20950-3330}{333\times {10}^3}\mathrm{kg}\\ & =& 52.91\times {10}^{-3}\mathrm{kg}\\ & \cong & 53\mathrm{g}\end{array}$

Therefore, we can say that ice and water reach thermal equilibrium at a temperature of 0^oC with 53 g of ice melted.

Now there is less than 53g ice present initially. All the ice melts, and the final temperature is above the melting point of ice. Using equation (i), the heat rejected by the water is given as:

${\mathrm{Q}}_1={\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}\left({\mathrm{T}}_{\mathrm{Wi}}-{\mathrm{T}}_{\mathrm{f}}\right)$ …(iii)

and using same equation (i), the heat absorbed by the ice and the water, it becomes when it melts is given as:

${\mathrm{Q}}_2={\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}\left(0-{\mathrm{T}}_{\mathrm{Ii}}\right)+{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{I}}\left({\mathrm{T}}_{\mathrm{f}}-0\right)+{\mathrm{M}}_{\mathrm{I}}{\mathrm{L}}_{\mathrm{f}}$ …(iv)

The first term is the energy required to raise the temperature of the ice to 0^oC, and the second term is the energy required to raise the temperature of the melted ice from 0^oC to${\mathrm{T}}_{\mathrm{f}}$and the third term is the energy required to melt all the ice. Since the two heats are equal, using equations (iii) and (iv) is given as:

${\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}\left({\mathrm{T}}_{\mathrm{Wi}}-{\mathrm{T}}_{\mathrm{f}}\right)={\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}\left(0-{\mathrm{T}}_{\mathrm{Ii}}\right)+{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{I}}\left({\mathrm{T}}_{\mathrm{f}}-0\right)+{\mathrm{M}}_{\mathrm{I}}{\mathrm{L}}_{\mathrm{f}}$

Therefore, the final temperature from the above equation is given as:

$\begin{array}{rcl}{\mathrm{T}}_{\mathrm{f}}& =& \frac{{\mathrm{C}}_{\mathrm{W}}{\mathrm{M}}_{\mathrm{W}}{\mathrm{T}}_{\mathrm{Wi}}+{\mathrm{C}}_{\mathrm{I}}{\mathrm{M}}_{\mathrm{I}}{\mathrm{T}}_{\mathrm{Ii}}-{\mathrm{M}}_{\mathrm{I}}{\mathrm{L}}_{\mathrm{f}}}{{\mathrm{C}}_{\mathrm{W}}\left({\mathrm{M}}_{\mathrm{W}}+{\mathrm{M}}_{\mathrm{I}}\right)}\\ & =& \frac{4190{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.200\mathrm{kg}\times {25}^{\mathrm{o}}\mathrm{C}+2220{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\times 0.100\mathrm{kg}\times \left(-{15}^{\mathrm{o}}\mathrm{C}\right)-0.100kg\times 333\times {10}^3{\displaystyle \frac{J}{kg}}}{4190{\displaystyle \frac{\mathrm{J}}{\mathrm{kg}.\mathrm{K}}}\left(0.200\mathrm{kg}+0.100\mathrm{kg}\right)}\\ & =& \frac{20950-3330-3330}{1257}{}^{\mathrm{o}}\mathrm{C}\\ & =& 2.5^{\mathrm{o}}\mathrm{C}\end{array}$

Hence, the final temperature is 2.5^oC.