Chapter 12.1, Problem 6PP

To determine

To Calculate: The final temperature of the mixture.

Answer to Problem 6PP

  $50°\text{C}$

Explanation of Solution

Given:

Mass of sample 1 of water, $m_1=2.00\times {10}^2\text{g}$

Initial temperature of sample 1 of water, $T_1=80.0°\text{C}$

Mass of sample 2 of water, $m_2=2.00\times {10}^2\text{g}$

Initial temperature of sample 2 of water, $T_2=10.0°\text{C}$

Formula Used:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

Calculations:

Specific heat capacity of water is $c=4.18\text{J/g}\cdot \text{°C}$

Let the final temperature of the mixture be T.

Heat lost by sample 1 = Heat gained by sample 2.

  $\begin{array}{l}-Q_1=Q_2\\ m_{1}c\left(T_1-T\right)=m_{2}c\left(T-T_2\right)\\ \because m_1=m_2\\ \left(T_1-T\right)=\left(T-T_2\right)\\ \left(80°\text{C}-T\right)=\left(T-20°\text{C}\right)\\ 100°\text{C}=2T\\ \Rightarrow T=50°\text{C}\end{array}$

Conclusion:

Thus, the final temperature of the mixture is $50°\text{C}$ .