Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> and <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> . The temperature of the hot reservoir of the first engine is <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> . and the temperature of the cold reservoir of the second engine is <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> . <math> <mi>T</mi> </math>  is temperature of the sink of first engine which is also the source for the second engine. How is <math> <mi>T</mi> </math> related to <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> </msub> </math> and <math> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msub> </math> , if both the engines perform equal amount of work?

W = Q 1 - Q 2 = Q 2 - Q 3 Q 2 = Q 1 + Q 3 2 2 = Q 1 Q 2 + Q 3 Q 2 2 = T 1 T + T 2 T T = T 1 + T 2 2

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Thermodynamic Process Physics Thermodynamics Physics Class 11th

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_{1}$ and $T_{2}$ . The temperature of the hot reservoir of the first engine is $T_{1}$ . and the temperature of the cold reservoir of the second engine is $T_{2}$ . $T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$ , if both the engines perform equal amount of work?

Option 1 -

$T = \sqrt[]{T_{1} T_{2}}$

Option 2 -

$T = \frac{T_{1} + T_{2}}{2}$

Option 3 -

$T = \frac{2 T_{1} T_{2}}{T_{1} + T_{2}}$

Option 4 -

$T = 0$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

W = Q_{1} - Q_{2} = Q_{2} - Q_{3}

$Q_{2} = \frac{Q_{1} + Q_{3}}{2}$

$2 = \frac{Q_{1}}{Q_{2}} + \frac{Q_{3}}{Q_{2}}$

$2 = \frac{T_{1}}{T} + \frac{T_{2}}{T}$

$T = \frac{T_{1} + T_{2}}{2}$

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alok kumar singh

Heat lost by steam $=$ Heat gained by water and calorimeter.

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N_a -> Avogadro’s Number

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