Chapter 4.7, Problem 3E

(A)

Interpretation Introduction

Interpretation:

The change in entropy of the heat reservoir.

Concept Introduction:

The change in entropy equation.

$\Delta S_{\text{reservoir}}=\frac{Q_{\text{reservoir}}}{T_{\text{reservoir}}}$

Here, heat is $Q$ and temperature is $T$.

(B)

Interpretation Introduction

Interpretation:

The change in entropy of the heat reservoir.

Concept Introduction:

The change in entropy equation.

$\Delta S_{\text{reservoir}}=\frac{Q_{\text{reservoir}}}{T_{\text{reservoir}}}$

Here, heat is $Q$ and temperature is $T$.

(C)

Interpretation Introduction

Interpretation:

The change in entropy of the heat reservoir.

Concept Introduction:

The change in entropy equation.

$\Delta S_{\text{reservoir}}=\frac{Q_{\text{reservoir}}}{T_{\text{reservoir}}}$

Here, heat is $Q$ and temperature is $T$.

Convert the heat in Joules $\left(Q\right)$.

$Q=\dot{W}t$

Here, power is $\dot{W}$ and time is $t$.