Chapter 12.1, Problem 7PP

To determine

The equilibrium temperature of water.

Answer to Problem 7PP

The equilibrium temperature of water is $56.54°\text{C}$

Explanation of Solution

Given:

A container has $1.00\times {10}^2 \text{g}$ of water at a temperature of $16.0°\text{C}$ . A piece of glass has mass $1.5\times {10}^2 \text{g}$ at a temperature of $70.0°\text{C}$ is placed in the container.

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. 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)

Calculation:

Here, the hot object is glass which has a mass of $m_A\text{=1}.\text{5}0\times \text{1}{0}^{\text{2}}\text{g}$ and has a specific heat of $C_A=840\text{ J/kg}\text{.K}$ . The water in the container has a mass of $m_B\text{=1}.00\times \text{1}{0}^{\text{2}}\text{g}$ and specific heat $C_B=4180\text{ J/kg}\cdot \text{K}$ .

The temperature of glass in Kelvin can be calculated as,

  $\begin{array}{c}{T}_A=70+273\\ =343\text{ K}\end{array}$

Similarly, the temperature of water in Kelvin is,

  $\begin{array}{c}{T}_A=16+273\\ =289\text{ K}\end{array}$

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

  $\begin{array}{c}{T}_f=\frac{1.5\times {10}^2\times 840\times 343+1.0\times {10}^2\times 4180\times 289}{1.5\times {10}^2\times 840+1.0\times {10}^2\times 4180}\\ =329.5\text{ K}\end{array}$

Thus, the equilibrium temperature in degree Celsius is,

  $\begin{array}{c}{T}_f=329.5-273\\ =56.54°\text{C}\end{array}$

Conclusion:

Thus, the equilibrium temperature of water is $56.54°\text{C}$ .