A reversible engine has an efficiency of $\frac{1}{4} .$ If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink:

Option 1 -

280°C

Option 2 -

382°C

Option 3 -

180.4°C

Option 4 -

174°C

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1 Answer

Answered by

Raj Pandey | Contributor-Level 9

2 months ago

Correct Option - 4

Detailed Solution:

$η = \frac{1}{4} o r 25 %$

$Δ T_{s i n k} = - 58 ° C (i e d e c r e a s e d)$

$η' = \frac{1}{2} o r 50 %$

T_{sin k} =?

As we know, the efficiency for a reversible engine is given by :

$η = 1 - \frac{T_{2}}{T_{1}} [T_{2} \to s i n k t e m p e r a t u r e]$

On comparing both, we get

$T_{2} = \frac{3}{4} T_{1} & (T_{2} - 58 ° C) = \frac{T_{1}}{2}$

$\Rightarrow T_{2} = 174 ° C$

<math> <mrow> <mi>η</mi> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mn>2</mn> <mn>5</mn> <mn>%</mn> </mrow> </math> <math> <mrow> <mi>Δ</mi> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mi>k</mi> </mrow> </msub> <mo>=</mo> <mo>−</mo> <mn>5</mn> <mn>8</mn> <mo>°</mo> <mi>C</mi> <mrow> <mo> (</mo> <mrow> <mi>i</mi> <mi>e</mi> <mtext> </mtext> <mtext> </mtext> <mi>d</mi> <mi>e</mi> <mi>c</mi> <mi>r</mi> <mi>e</mi> <mi>a</mi> <mi>s</mi> <mi>e</mi> <mi>d</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> <math> <mrow> <mi>η</mi> <mo>'</mo> <mo>=</mo> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>o</mi> <mi>r</mi> <mtext> </mtext> <mtext> </mtext> <mtext> </mtext> <mn>5</mn> <mn>0</mn> <mn>%</mn> </mrow> </math> Tsin k =? As we know, the efficiency for a reversible engine is given by : <math> <mrow> <mi>η</mi> <mo>=</mo> <mn>1</mn> <mo>−</mo> <mfrac> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mrow> <mo> [</mo> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>→</mo> <mi>s</mi> <mi>i</mi> <mi>n</mi> <mtext> </mtext> <mtext> </mtext> <mi>k</mi> <mtext> </mtext> <mtext> </mtext> <mi>t</mi> <mi>e</mi> <mi>m</mi> <mi>p</mi> <mi>e</mi> <mi>r</mi> <mi>a</mi> <mi>t</mi> <mi>u</mi> <mi>r</mi> <mi>e</mi> </mrow> <mo>]</mo> </mrow> </mrow> </math> On comparing both, we get <math> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mfrac> <mrow> <mn>3</mn> </mrow> <mrow> <mn>4</mn> </mrow> </mfrac> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>&</mo> <mrow> <mo> (</mo> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>−</mo> <mn>5</mn> <mn>8</mn> <mo>°</mo> <mi>C</mi> </mrow> <mo>)</mo> </mrow> <mo>=</mo> <mfrac> <mrow> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> </mrow> </math> <math> <mrow> <mo>⇒</mo> <msub> <mrow> <mi>T</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <mn>1</mn> <mn>7</mn> <mn>4</mn> <mo>°</mo> <mi>C</mi> </mrow> </math>

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A reversible engine has an efficiency of $\frac{1}{4} .$ If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink:

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A reversible engine has an efficiency of $\frac{1}{4} .$ If the temperature of the sink is reduced by 58°C, its efficiency becomes double. Calculate the temperature of the sink: