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Expert-verified Found in: Page 130 ### An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279 # In an absorption refrigerator, the energy driving the process is supplied not as work, but as heat from a gas flame. (Such refrigerators commonly use propane as fuel, and are used in locations where electricity is unavailable.* ) Let us define the following symbols, all taken to be positive by definition:Qf= heat input from flameQc= heat extracted from inside refrigeratorQr= waste heat expelled to roomTf= temperature of flameTc= temperature inside refrigeratorTr= room temperature(a) Explain why the "coefficient of performance" (COP) for an absorption refrigerator should be defined as Qc / Qf.(b) What relation among Qf, Qc, and Qr is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Tf, Tc, and Tr alone.

a)$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

b) $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$, COP can be greater than 1.

c) The value is $\mathrm{COP}=\frac{{T}_{\mathrm{c}}\left({T}_{\mathrm{f}}-{T}_{\mathrm{r}}\right)}{{T}_{\mathrm{f}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{c}}\right)}$

See the step by step solution

## Part(a): Step 1: Given Information

Qf= heat input from flameQc= heat extracted from inside refrigerator

Explain why COP for an absorption refrigerator should be defined as ${Q}_{\mathrm{c}}/{Q}_{\mathrm{f}}$

## Part(a): Step 2: Explanation

In an absorption refrigerator electrical energy is transformed into the heat from a gas flame.
Since the heat extracted is the benefit and the fuel is the cost in operating the refrigerator therefore the coefficient of performance isSo input energy is Qf and output is Qc so COP is written as$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

## Part(b): Step 1: Given information

Qf= heat input from flameQc= heat extracted from inside refrigeratorQr= waste heat expelled to room

and $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}$

Find relationship among Qf, Qc, and Qr ?

Will energy conservation permit the COP to be greater than 1 ?

## Part (b): Step 2 : Explanation

In an absorption refrigerator heat from a gas flame is transformed into electrical energy

We have $\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{{Q}_{\mathrm{f}}}..........................\left(1\right)$

Rewrite the expression to include the waste heat QR expelled to room, we get

${Q}_{\mathrm{r}}={Q}_{\mathrm{f}}+{Q}_{\mathrm{c}}$

Rearrange the equation and Substitute this in the equation (1)

$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$

Simplify this by dividing by Qc

role="math" localid="1647272341714" $\text{COP}=\frac{1}{\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}-1}.............................\left(2\right)$

From the equation(2) ${Q}_{\mathrm{r}}/{Q}_{\mathrm{c}}$is greater than 2, so COP can be greater than 1.

## Part(C): Step 1:Given

Tf= temperature of flameTc= temperature inside refrigeratorTr= room temperature

$\mathrm{COP}=\frac{{Q}_{\mathrm{c}}}{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}$

Find the upper limit of COP

## Part(c): Step 2: Explanation

In an absorption refrigerator exchanges the electrical energy with the heat energy.

Use the second law of thermodynamics write the expression of the condition

$\frac{{Q}_{\mathrm{f}}}{{T}_{\mathrm{f}}}+\frac{{Q}_{\mathrm{c}}}{{T}_{\mathrm{c}}}=\frac{{Q}_{\mathrm{r}}}{{T}_{\mathrm{r}}}$

Substitute ${Q}_{\mathrm{f}}=\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)$in the above equation

$\frac{\left({Q}_{\mathrm{r}}-{Q}_{\mathrm{c}}\right)}{{T}_{\mathrm{f}}}+\frac{{Q}_{\mathrm{c}}}{{T}_{\mathrm{c}}}=\frac{{Q}_{\mathrm{r}}}{{T}_{\mathrm{r}}}$

Rearrange the above equation,

$\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}=\frac{{T}_{\mathrm{r}}\left({T}_{\mathrm{c}}-{T}_{\mathrm{f}}\right)}{{T}_{\mathrm{c}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{f}}\right)}$

Substitute $\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}=\frac{{T}_{\mathrm{r}}\left({T}_{\mathrm{c}}-{T}_{\mathrm{f}}\right)}{{T}_{\mathrm{c}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{f}}\right)}intheCOPformula\text{COP}=\frac{1}{\frac{{Q}_{\mathrm{r}}}{{Q}_{\mathrm{c}}}-1}$

$\mathrm{COP}=\frac{{T}_{\mathrm{c}}\left({T}_{\mathrm{f}}-{T}_{\mathrm{r}}\right)}{{T}_{\mathrm{f}}\left({T}_{\mathrm{r}}-{T}_{\mathrm{c}}\right)}$ ### Want to see more solutions like these? 