Question

Two Carnot engines A and B are operated in series. Engine A receives heat from a reservoir at 600 K and rejects heat to a reservoir heat to reservoir at temperature T. Engine B receives heat rejected by engine A and in turn rejects it to a reservoir at 100 K. If the efficiencies of the two engines A and B are represented by ${{\eta }_{A}}$ and ${{\eta }_{B}}$ respectively, then what is the value of $\dfrac{{{\eta }_{B}}}{{{\eta }_{A}}}$.
A. $\dfrac{12}{7}$
B. $\dfrac{12}{5}$
C. $\dfrac{5}{12}$
D. $\dfrac{7}{12}$

Answer 1

Hint: To solve this question, we should have prior knowledge about the formula of efficiency. The question is based on the correct substitution of values in the efficiency equation.

Complete step by step answer:
 First let us consider the two efficiencies that are provided for systems or rather the engines A and B.
Beginning with engine A, we know the efficiency of engine A is represented as ${{\eta }_{A}}$.
So, we can write that efficiency is given by ${{\eta }_{A}}\,\text{= }1-\dfrac{\text{T}}{\text{600}}$ (considering the efficiency equation of engine A)
Moving on to engine B. We know that the efficiency of engine B is represented as ${{\eta }_{B}}$.
So, we can write that efficiency is given by ${{\eta }_{B}}\text{ = }1-\dfrac{\text{100}}{\text{T}}$ (considering the efficiency equation of engine B)
Once we know the efficiency of both the engines, we must find the amount of work done by both the engines, that is engine A and engine B.
The work produced by engine A and B is same so intermediate temperature $\text{T=}\dfrac{{{\text{T}}_{\text{1}}}\text{+}{{\text{T}}_{\text{3}}}}{\text{2}}$
In the given question,
${{\text{T}}_{\text{1}}}$ represents the time taken by engine A for the heat input.
$\text{T}$ represents the time taken for the engine A to give the heat as output to the engine B. We can also say that the heat from engine A is received by engine B, at a temperature of $\text{T}$.
${{\text{T}}_{\text{3}}}$represents the temperature at which the engine B gives out the heat as output to the sink.
We can say that ,
${{\text{T}}_{\text{1}}}\text{= 600K}$
${{\text{T}}_{\text{3}}}\text{=100K}$
So, $\text{T= 350 K}$
Now putting the values in the efficiency equation of both the engines.
${{\text{ }\!\!\eta\!\!\text{ }}_{\text{A}}}\text{=1-}\dfrac{\text{350}}{\text{600}}$and
${{\text{ }\!\!\eta\!\!\text{ }}_{\text{B}}}\text{=1-}\dfrac{\text{100}}{\text{350}}$
Now, at the end we are required to find $\dfrac{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{B}}}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{A}}}}$.
So, $\dfrac{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{B}}}}{{{\text{ }\!\!\eta\!\!\text{ }}_{\text{A}}}}$ gives us the value of $\dfrac{12}{7}$
Therefore, the correct answer is Option A.
Note: In physics, efficiency is defined as the comparison between the output energy and the input energy of a given system. Efficiency can also be defined as the percentage ratio of the output energy to the input energy, for a given equation. The main purpose of the efficiency equation is to represent the energy of a given system, in the form of heat and power.
The significance of efficiency is that; it gives us an assumption of the peak level of performance involving the least amount of inputs, to obtain the highest possible output. The advantage of knowing the efficiency is that it reduces the wastage of resources, and helps in achieving the input, within the desired time.