Question

Part A A heat engine is designed to do work. This is possible only if certain relationships between the heats and temp...

Part A
A heat engine is designed to do work. This is possible only if certain relationships between the heats and temperatures at the input and output hold true. Which of the following sets of statements must apply for the heat engine to do work?

A) Qh < Qc and Th < Tc
B) Qh > Qc and Th < Tc
C) Qh < Qc and Th > Tc
D) Qh > Qc and Th > Tc

Part B
Find the work W done by the "ideal" heat engine.
Express W in terms of Qh and Qc
W=

Part C
The thermal efficiency e of a heat engine is defined as follows: e = W/Qh
Express the efficiency in terms of Qh and Qc.
e=
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Answer #1
Concepts and reason

The given problem can be solved by using the concept of heat engine, the expression of work done by the engine and efficiency of the engine.

Initially calculate the work done by the ideal heat engine and then the efficiency of heat engine will be calculated.

Fundamentals

The expression for the work done can be expressed as follows,

W=QhQcW = {Q_h} - {Q_c}

Here, Qh{Q_h} is the heat flowing into the heat engine and Qc{Q_c} is the heat wasted into the sink.

The expression for efficiency of the engine can be expressed as follows,

η=WQh\eta = \frac{W}{{{Q_h}}}

Here, W is the work done by the heat engine and Qh is the quantity of heat

Or it can be expressed as follows,

η=QhQcQh\eta = \frac{{{Q_h} - {Q_c}}}{{{Q_h}}}

(A)

Consider the heat engine diagram as follows,

(T.) Hot
Reservor
Qu
Heat Engine
T = Temperature (K)
Q = Heat (J)
W = Work (J)
h = hot
c= cold
Heat
Engine
Efficiency
W=Q-Q
Q

The relation between heat and temperature at input and output is as, Qh>Qc{Q_h} > {Q_c} and Th>Tc{T_h} > {T_c} .

(B)

The expression for the work done by the engine can be expressed as follows,

W=QhQcW = {Q_h} - {Q_c}

(C)

The expression for efficiency of the engine can be expressed as follows,

η=WQh\eta = \frac{W}{{{Q_h}}}

Substitute QhQc{Q_h} - {Q_c} for WW in the expression of η\eta .

η=QhQcQh=1QcQh\begin{array}{c}\\\eta = \frac{{{Q_h} - {Q_c}}}{{{Q_h}}}\\\\ = 1 - \frac{{{Q_c}}}{{{Q_h}}}\\\end{array}

Ans: Part A

The relation between heat and temperature at input and output is as, Qh>Qc{Q_h} > {Q_c} and Th>Tc{T_h} > {T_c} .

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