Question

An engine operating between \[{\rm{15}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]and \[{\rm{2}}{{\rm{5}}^{\rm{0}}}{\rm{C}}\] takes \[{\rm{500}}\,{\rm{J}}\] heat from a higher temperature reservoir. If there are no frictional losses, then work done by the engine is:
A. \[147.7\,{\rm{J}}\]
B. \[157.75\,{\rm{J}}\]
C. \[165.85\,{\rm{J}}\]
D. \[169.95\,{\rm{J}}\]

Answer 1

Hint: The fraction of the heat absorbed by a machine that it can transform into work is known as the efficiency of the machine. The machine used for the conversion of heat into work is called the heat engine.

Formula used
Work done can be calculated by using the relationship as shown below.
\[\dfrac{{\rm{W}}}{{\rm{q}}} = \dfrac{{{{\rm{T}}_{\rm{2}}} - {{\rm{T}}_1}}}{{{{\rm{T}}_{\rm{2}}}}}\]
where, \[{\rm{W}}\]= work done
\[{\rm{q}}\]= heat
\[{{\rm{T}}_1}\]= temperature of the sink
\[{{\rm{T}}_{\rm{2}}}\]= temperature of the source

Complete Step by Step Solution:
In order to convert the heat into work, the heat engine absorbs heat from a heat reservoir at a higher temperature known as the source, where it converts a part of heat into work and returns the remainder to the heat reservoir at a lower temperature known as the sink.
As per the given data,
Initial temperature, \[{{\rm{T}}_1}{\rm{ = 2}}{{\rm{5}}^{\rm{0}}}{\rm{C}}\]
Final temperature, \[{{\rm{T}}_2}{\rm{ = 15}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\]
Heat, \[{\rm{q = 500}}\,{\rm{J}}\]
Convert the given temperatures from degree Celsius to Kelvin by using the relationship, \[{\rm{K}}{{\rm{ = }}^{\rm{0}}}{\rm{C}} + 273\]as shown below.
So, the sink temperature and source temperature will become as:
\[{{\rm{T}}_1}{\rm{ = 2}}{{\rm{5}}^{\rm{0}}}{\rm{C = (25 + 273)K = 298K}}\]
\[{{\rm{T}}_2}{\rm{ = 15}}{{\rm{0}}^{\rm{0}}}{\rm{C = (150 + 273)K = 423K}}\]
Find the work done by the heat engine by using the relationship as given below.
\[\dfrac{{\rm{W}}}{{\rm{q}}} = \dfrac{{{{\rm{T}}_{\rm{2}}} - {{\rm{T}}_1}}}{{{{\rm{T}}_{\rm{2}}}}}\]
Rearrange the above formula and solve for \[{\rm{W}}\].
\[{\rm{W}} = {\rm{q}}(\dfrac{{{{\rm{T}}_{\rm{2}}} - {{\rm{T}}_1}}}{{{{\rm{T}}_{\rm{2}}}}})\]
Substituting the values given, we get as:
\[\begin{array}{c}{\rm{W}} = {\rm{q}}(\dfrac{{{{\rm{T}}_{\rm{2}}} - {{\rm{T}}_1}}}{{{{\rm{T}}_{\rm{2}}}}})\\ = 500\,{\rm{J}}(\dfrac{{423{\rm{K}} - 298{\rm{K}}}}{{423{\rm{K}}}})\\ = 500\,{\rm{J}} \times (\dfrac{{125{\rm{K}}}}{{423{\rm{K}}}})\\ = 147.7\,{\rm{J}}\end{array}\]
Hence, the work done by the heat engine is calculated to be as \[147.7\,{\rm{J}}\]
Therefore, option A is correct.

Note: The relation \[\dfrac{{\rm{W}}}{{\rm{q}}} = \dfrac{{{{\rm{T}}_{\rm{2}}} - {{\rm{T}}_1}}}{{{{\rm{T}}_{\rm{2}}}}}\]gives the efficiency of the Carnot cycle or engine. It is clear that the efficiency of the reversible heat engine depends only upon the temperatures of the source and sink and is independent of the nature of the working substance.