Chapter 15, Problem 57GP

(a)

To determine

The final temperature of the water.

Answer to Problem 57GP

The final temperature of the water is $44.55°\text{C}$ .

Explanation of Solution

Given info:

The mass of the insulated aluminum cup is, $m_a=120\text{g}$ .

The mass of the water is, $m_w=140\text{g}$ .

The temperature of the aluminum cup is, $T_a=15°\text{C}$ .

The temperature of the water is, $T_w=50°\text{C}$ .

Formula Used:

The expression to calculate the final temperature of the water is,

  $T_f=\frac{m_a{c}_a{T}_a+m_w{c}_w{T}_w}{m_a{c}_a+m_w{c}_w}$

Here,

• $c_a$ is the specific heat of the Aluminum and its value is $0.9\text{J}/\text{g}\cdot °\text{C}$ .
• $c_w$ is the specific heat of the water and its value is $4.186\text{J}/\text{g}\cdot °\text{C}$ .

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{T}_f=\frac{\left(120\text{g}\right)\left(0.9\text{J}/\text{g}\cdot °\text{C}\right)\left(15°\text{C}\right)+\left(140\text{g}\right)\left(4.186\text{J}/\text{g}\cdot °\text{C}\right)\left(50°\text{C}\right)}{\left(120\text{g}\right)\left(0.9\text{J}/\text{g}\cdot °\text{C}\right)+\left(140\text{g}\right)\left(4.186\text{J}/\text{g}\cdot °\text{C}\right)}\\ =44.55°\text{C}\end{array}$

Conclusion:

Thus, the final temperature of the water is $44.55°\text{C}$ .

(b)

To determine

The total change in entropy.

Answer to Problem 57GP

The change in entropy is $149.05\text{J}/°\text{C}$ .

Explanation of Solution

Formula Used:

The expression to calculate the amount of heat lost is,

  $Q=m_w{c}_w\left(T_w-T_f\right)$

The expression to calculate the change in entropy is,

  $\Delta s=Q\left(\frac{1}{T_a}-\frac{1}{T_w}\right)$

Calculation:

Substitute the values in the above expression.

  $\begin{array}{c}Q=\left(140\text{g}\right)\left(4.186\text{J}/\text{g}\cdot °\text{C}\right)\left(50°\text{C}-44.55°\text{C}\right)\\ =3193.92\text{J}\end{array}$

Substitute the values in the above expression.

  $\begin{array}{c}\Delta s=\left(3193.92\text{J}\right)\left(\frac{1}{15°\text{C}}-\frac{1}{50°\text{C}}\right)\\ =149.05\text{J}/°\text{C}\end{array}$

Conclusion:

Thus, the change in entropy is $149.05\text{J}/°\text{C}$ .