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Standard XII

Physics

2nd Equation of Motion

Two carnot en...

Question

Two carnot engine are connected in series and output work of both engine is same. If temperature of source of $1^{s t}$ engine is 600 K and temperature of sink of $2^{n d}$ engine is 400 K. The temperature of junction will be

300 K

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400 K

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500 K

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600 K

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Solution

The correct option is C 500 K
$W = Q_{1} - Q_{2}$

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

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

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

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

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

$T_{2} = \frac{T_{1} + T_{3}}{2} = 500 K$

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Two carnot engine are connected in series and output work of both engine is same. If temperature of source of 1st engine is 600 K and temperature of sink of 2nd engine is 400 K. The temperature of junction will be

The correct option is C 500 KW=Q1−Q2 W=Q2−Q3 Q1−Q2=Q2−Q3 Q1+Q3=2Q2 Q1Q2+Q3Q2=2 T1T2+T3T2=2 T2=T1+T32=500 K

Two carnot engine are connected in series and output work of both engine is same. If temperature of source of $1^{s t}$ engine is 600 K and temperature of sink of $2^{n d}$ engine is 400 K. The temperature of junction will be

The correct option is C 500 K
$W = Q_{1} - Q_{2}$

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

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

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

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

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

$T_{2} = \frac{T_{1} + T_{3}}{2} = 500 K$