Chapter 12.1, Problem 8PP

To determine

The final temperature of the mixture.

Answer to Problem 8PP

The final temperature of the mixture is $42.1°\text{C}$

Explanation of Solution

Given:

A water sample of $4.0\times {10}^2 \text{g}$ at a temperature of $15.0°\text{C}$ is mixed with water of $4.0\times {10}^2 \text{g}$ at a temperature of $85.0°\text{C}$ . Then $4.0\times {10}^2 \text{g}$ of methanol at temperature $15.0°\text{C}$ is added to the mixture at equilibrium. There is no loss of energy to container or surroundings.

Formula Used:

When a hot object at a temperature $T_A$ (in Kelvin) is immersed in cooler water at initial temperature $T_B$ (Kelvin) , the temperature of water increases and reaches an equilibrium value. This is due to the exchange of heat energy between the object and water. It is also being assumed that there is no loss of heat to surroundings or the container.

Now, if the mass of the object and water is $m_A$ and $m_B$ respectively, with specific heat $C_A$ and $C_B$ , the equilibrium temperature is given by the expression,

  $T_f=\frac{m_A{C}_A{T}_A+m_B{C}_B{T}_B}{m_A{C}_A+m_B{C}_B}$ ..... (1)

Now, a third substance is added to the mixture after reaching equilibrium. This can be treated in the same way as the two-substance case, with one of the substance being the mixture itself and the other being the third substance.

Calculation:

Here, firstly two components(which is water) are mixed together. Both the hot water and the cold water are having the same mass of $m_A=m_B\text{=4}.0\times \text{1}{0}^{\text{2}}\text{g}$ . The hot water is kept at a temperature $85.0°\text{C}$ . The cold water is at a temperature $15.0°\text{C}$ . The specific heat is $C_A=C_B=4180\text{ J/kg}\cdot \text{K}$ .

The temperature of hot water in Kelvin can be calculated as,

  $\begin{array}{c}{T}_A=85+273\\ =358\text{ K}\end{array}$

Similarly, the temperature of cold water in Kelvin is,

  $\begin{array}{c}{T}_A=15+273\\ =288\text{ K}\end{array}$

Substituting the given parameters in equation (1), equilibrium temperature is obtained as,

  $\begin{array}{c}{T}_f=\frac{4\times {10}^2\times 4180\times 358+4.0\times {10}^2\times 4180\times 288}{4\times {10}^2\times 4180+4.0\times {10}^2\times 4180}\\ =323\text{ K}\end{array}$

Now, $\text{4}.0\times \text{1}{0}^{\text{2}}\text{g}$ of methanol at a temperature $15.0°\text{C}$ is added to the above mixture at 323K. The specific heat of methanol is $C_M=2450\text{ J/kg}\cdot \text{K}$

The temperature of methanol in Kelvin is,

  $\begin{array}{c}{T}_M=15+273\\ =288\text{ K}\end{array}$

In the new mixture, the other component is water with a mass $\text{8}.0\times \text{1}{0}^{\text{2}}\text{g}$

Therefore, considering (1) again for the new mixture, the equilibrium temperature can be obtained as,

  $\begin{array}{c}{T}_f=\frac{8\times {10}^2\times 4180\times 323+4.0\times {10}^2\times 2450\times 288}{8\times {10}^2\times 4180+4.0\times {10}^2\times 2450}\\ =315.06\text{ K}\\ \text{=315}\text{.1 K}\end{array}$

Thus, the equilibrium temperature of the final mixture in degree Celsius is,

  $\begin{array}{c}{T}_f=315.1-273\\ =42.1°\text{C}\end{array}$

Conclusion:

Thus, the equilibrium temperature of final mixture is $42.1°\text{C}$ .