Physics, 10.09.2020 01:01 trin520

A reversible heat engine, operating in a cycle, withdraws thermal energy from a high-temperature reservoir (the temperature of which consequently decreases), performs work w, and rejects thermal energy into a low-temperature reservoir (the temperature of which consequently increases). The two reservoirs are, initially, at the temperatures T1 and T2 and have constant heat capacities C1 and C2, respectively. Calculate the final temperature of the system and the maximum amount of work which can be obtained from the engine.

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Answer from: payshencec21

The final temperature is $\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}$

and the maximum amount of workdone is $C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}$ .

Explanation:

Assume that R_1 is the reservior having temperature T_1 K and heat capicity $C_1 \frac JK$ and R_2 is the reservior having temperature T_2 and heat capicity $C_2 \frac JK$ .

The work will be extracted till that both the reservior reach the thermal equilibrium. Let the final temperature of both the reservior is T_f .

Let total Q_1 heat is extracted by the heat engine from the reservior R_1 and its temperature decreases from T_1 to T_f and Q_2 heat is rejected by the heat engine to the reservior R_2 and its temperature decreases from T_2 to T_f .

So, The maximum amount of work done,

$w= Q_1 - Q_2\; \cdots (i)$

Now, as the heat engine is reversible, so change is entropy for the universe is 0, which means sum of change in entropy for the ststem as well as surrounding is 0.

As shown in figure, the system is the reversible engine, so, change is entropy for the system is 0. Hence, change in entropy for the the surrounding is 0.

As temperature of R_1 is changing fron T_1 to T_f , so, change in entropy of surrounding due to transfer of Q_1 is $C_1 \ln \frac {T_f}{T_1}$ .

Similarly, change in entropy of surrounding due to transfer of Q_2 is $C_2 \ln \frac {T_f}{T_2}$ .

As the net change in entropy of the surrounding is 0.

$\Rightarrow C_1 \ln \frac {T_f}{T_1}+C_2 \ln \frac {T_f}{T_2}=0$

$\Rightarrow \ln \left( \frac {T_f}{T_1} \right)^{C_1}+ \ln \left( \frac {T_f}{T_2}\right)^{C_2}=0$

$\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}=- \ln \left( \frac {T_f}{T_2}\right)^{C_2}$

$\Rightarrow \ln \left(\frac {T_f}{T_1}\right)^{C_1}= \ln \left( \frac {T_2}{T_f}\right)^{C_2}$

$\Rightarrow \left( \frac {T_f}{T_1}\right)^{C_1}=\left( \frac {T_2}{T_f}\right)^{C_2}$ [taking anti-log both the sides]

$\Rightarrow T_f^{(C_1 +C_2)}=T_1^{C_1}+T_2^{C_2}$

$\Rightarrow T_f=\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}\; \cdots (ii)$

This is the required final temperature.

Now, from equarion (i), the maximum amount of work done is

w= Q_1 - Q_2

As $Q=C\Delta T$

$\Rightarrow w=C_1(T_1-T_f)-C_2(T_f-T_2)$

$\Rightarrow w=C_1T_1 + C_2T_2-(C_1+C_2)T_f$

From equation (ii) ,

$w=C_1T_1 + C_2T_2-(C_1+C_2)\left(T_1^{C_1}+T_2^{C_2}\right)^{\frac {1}{C_1+C_2}}$

This is the required maximum workdone.

A reversible heat engine, operating in a cycle, withdraws thermal energy from a high-temperature res

Answer from: Quest

what is the question because that is not a question.

Answer from: Quest

the design of a ski jumping suit involves a trade-off between giving the skiers lift to get higher jumps and ensuring the skier does not face a lot of drag before the jump itself. an expansive suit may to lift the skier higher, but at the same time, it will also increase the air resistance. higher air resistance as the skier goes down the slope limits the speed of the jump. a big suit also increases the risk of losing balance or control, especially if there's a strong gust of wind. while technologically enhanced suits a lot, clothing is not the only factor in a ski jumper’s success. gathering speed while skiing down the slope, ensuring good speed on the ramp, and maintaining good body posture and form in the air all contribute to a ski jumper's success.

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