Solve the problem.
-A heat engine is a device that converts thermal energy into other forms. The thermal efficiency, e, of a heat engin defined by
$$\mathrm { e } = \frac { \mathrm { Q } _ { \mathrm { h } } - \mathrm { Q } _ { \mathrm { C } } } { \mathrm { Q } _ { \mathrm { h } } } \text {, }$$
where $\mathrm { Q } _ { h }$ is the heat absorbed in one cycle and $\mathrm { Q } _ { \mathrm { C } }$, the heat released into a reservoir in one cycle, is a constant. Find $\frac { d ^ { 2 } e } { d h ^ { 2 } }$
A) $\frac { \mathrm { d } ^ { 2 } \mathrm { e } } { \mathrm { dQ } _ { \mathrm { h } ^ { 2 } } } = \frac { \mathrm { Q } _ { \mathrm { C } } } { \mathrm { Q } _ { \mathrm { h } ^ { 2 } } }$
B) $\frac { d ^ { 2 } e } { d Q _ { h } ^ { 2 } } = \frac { - 2 Q _ { C } } { Q _ { h } { } ^ { 3 } }$
C) $\frac { d ^ { 2 } e } { d _ { h ^ { 2 } } ^ { 2 } } = \frac { Q _ { c } } { Q _ { h ^ { 3 } } }$
D) $\frac { \mathrm { d } ^ { 2 } \mathrm { e } } { \mathrm { dQ } _ { \mathrm { h } } ^ { 2 } } = \frac { - \mathrm { Q } _ { \mathrm { C } } } { 2 \mathrm { Qh } ^ { 2 } }$

Question

Quizplus · Accepted Answer

First, let's find the first derivative of e with respect to Q_h:
\begin{align*} \frac{de}{dQ_h} &= \frac{d}{dQ_h}\left(\frac{Q_h-Q_C}{Q_h}ight) \ &= \frac{d}{dQ_h}\left(1-\frac{Q_C}{Q_h}ight) \ &= \frac{Q_C}{Q_h^2} \end{align*}
Now, let's find the second derivative of e with respect to Q_h:
\begin{align*} \frac{d^2e}{dQ_h^2} &= \frac{d}{dQ_h}\left(\frac{Q_C}{Q_h^2}ight) \ &= -\frac{2Q_C}{Q_h^3} \end{align*}
Therefore, the answer is B.

Solve the Problem e=Qh−QCQh, \mathrm { e } = \frac { \mathrm { Q } _ { \mathrm { h } } - \mathrm { Q } _ { \mathrm { C } } } { \mathrm { Q } _ { \mathrm { h } } } \text {, }e=Qh​Qh​−QC​​,

Solve the Problem $\mathrm { e } = \frac { \mathrm { Q } _ { \mathrm { h } } - \mathrm { Q } _ { \mathrm { C } } } { \mathrm { Q } _ { \mathrm { h } } } \text {, }$