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# A reversible engine takes heat from a reservoir at ${527^ \circ }C$ and gives out to the sink at ${127^ \circ }C$. The engine is required to perform useful mechanical work at the rate of $750watt$. How many calories per second must it take from the reservoir?(A) $357.1cal/\sec$(B) $257.1cal/\sec$(C) $157.1cal/\sec$(D) $57.1cal/\sec$

Last updated date: 13th Jun 2024
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Hint: A Carnot engine operating between two given temperatures has the greatest possible efficiency of any heat engine operating between these two temperatures. All engines employing only reversible processes have this same maximum efficiency when operating between the same given temperatures. Work done by the engine is the difference between the amount of heat absorbed and the amount of heat released.
Formula Used: The formulae used in the solution are given here.
The efficiency of the engine is given by $\varepsilon = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$ where ${T_1}$ is the temperature of the source and ${T_2}$ is the sink temperature.
The efficiency of the engine is given by $\varepsilon = \dfrac{W}{{{Q_1}}}$ where ${Q_1}$ is the number of calories per second taken from the reservoir.

Complete Step by Step Solution: It has been given that a reversible engine takes heat from a reservoir at ${527^ \circ }C$ and gives out to the sink at ${127^ \circ }C$. The engine is required to perform useful mechanical work at the rate of $750watt$.
We know that a Carnot heat engine is a theoretical engine that operates on a reversible Carnot cycle. It has maximum efficiency that a heat engine can possess. It involves four processes: isothermal expansion, adiabatic expansion, isothermal compression and adiabatic compression.
Carnot’s theorem states that the heat engines that are working between two heat reservoirs are less efficient than the Carnot heat engine that is operating between the same reservoirs.
Irrespective of the operation details, every Carnot engine is efficient between two heat reservoirs.
Maximum efficiency is given as, $\varepsilon = \dfrac{{{T_1} - {T_2}}}{{{T_1}}}$ where ${T_1}$ is the temperature of the source and ${T_2}$ is the sink temperature.
Given that, ${T_1} = 527 + 273 = 800K$ and ${T_2} = 127 + 273 = 400K$.
Thus, the efficiency is, $\varepsilon = \dfrac{{800 - 400}}{{800}} = \dfrac{1}{2}$.
When expressed as percentage efficiency is 50%.
Now we know that, the efficiency is also given by, $\varepsilon = \dfrac{W}{{{Q_1}}}$ where ${Q_1}$ is the number of calories per second taken from the reservoir.
Thus, ${Q_1} = \dfrac{W}{\varepsilon }$ where $W$ is the work done.
Now, work done $W = \dfrac{{750}}{{1/2}} = 1500J$. We know that $1cal = 4.2J$.
The work done is, $W = \dfrac{{1500}}{{4.2}} = 357.1cal/\sec$.

Thus the correct answer is Option A.

Note: The efficiency of the engine is also given by $\varepsilon = \dfrac{{{Q_1} - {Q_2}}}{{{Q_1}}}$ where ${Q_2}$ is the amount of heat released and ${Q_1}$ is the amount of heat absorbed. Work done is not required to solve this question, it must not create confusion.