Chapter 15, Problem 1P

(a)

To determine

The W in the process.

Answer to Problem 1P

  $+1.6\times {10}^4\text{J}$

Explanation of Solution

Given:

The work done by the student, $W=1.6\times {10}^4\text{J}$

The decrease in the internal energy of the student, $\Delta U=-4.2\times {10}^4\text{J}$

Calculation:

Since the student does the work, hence the “ W ” of the process will be positive. Hence,

  $W=+1.6\times {10}^4\text{J}$

Conclusion:

The W in the processis $+1.6\times {10}^4\text{J}$ .

(b)

To determine

The $\Delta U$ in the process.

Answer to Problem 1P

  $\Delta U=-4.2\times {10}^4\text{J}$

Explanation of Solution

Given:

The work done by the student, $W=1.6\times {10}^4\text{J}$

The decrease in the internal energy of the student, $\Delta U=-4.2\times {10}^4\text{J}$

Calculation:

Since the internal energy of the student decreases. Hence, it can be written as,

  $\Delta U=-4.2\times {10}^4\text{J}$

Conclusion:

The $\Delta U$ in the process is $-4.2\times {10}^4\text{J}$ .

(c)

To determine

The $Q$ in the process.

Answer to Problem 1P

  $-2.6\times {10}^4\text{J}$

Explanation of Solution

Given:

The work done by the student, $W=1.6\times {10}^4\text{J}$

The decrease in the internal energy of the student, $\Delta U=-4.2\times {10}^4\text{J}$

Formula used:

First law of thermodynamics,

  $Q=\Delta U+W$

Where $Q$ is the heat gained by the system, $\Delta W$ is the work done by the system on the surroundings and $\Delta U$ is the change in internal energy of the system.

Calculation:

Using given values in above equation, we get,

  $\begin{array}{l}Q=\left(-4.2\times {10}^4\right)+\left(1.6\times {10}^4\right)\\ Q=-2.6\times {10}^4\text{J}\end{array}$

Conclusion:

The $Q$ in the process is $-2.6\times {10}^4\text{J}$ .