Question

The thermal efficiency of a heat engine for which the work output is 2500J/cycle and the heat input is 10000J/cycle, is:
A. 75%
B. 40%
C. 25%
D. 0.25%

Answer 1

Hint:Thermal efficiency for a heat engine is defined as the ratio of work delivered by the engine to the amount of heat supplied to the engine. The value of thermal efficiency is always less than 100 % as some amount of heat is lost while doing work.

Complete Step by Step Answer:
Given:
The work output of the engine is \[W = 2500\,J/cycle\] .
The heat input to the engine is $Q = 10000J/cycle$ .

Express the relation for the efficiency of a thermal heat engine.
$\eta = \dfrac{W}{Q} \times 100$
Here $\eta $ is the efficiency of the engine.
Substitute 2500 J/cycle for W and 10000 J/cycle for Q to find the value of $\eta $.
\[\begin{array}{l}
\eta = \dfrac{{2500J/cycle}}{{10000J/cycle}} \times 100\\
\eta = \dfrac{{1J/cycle}}{{4J/cycle}} \times 100\\
\eta = 25\%
\end{array}\]

Here the efficiency is 25% which means that out of the total energy supplied, only the 25% of that energy is converted into work. Therefore, the correct option is (C).

Efficiency is the ratio between energy given and energy output. For an instant, the efficiency of a truck engine is typically only 25%. The losses are mainly caused due to waste heat and friction involved. Even the normal lever is not 100% efficient since there are losses associated due to friction at the fulcrum.

Note:There is always a basic trade-off and confusion between the efficiency and mechanical advantage. Mechanical advantage depicts the multiply of force/torque by mechanical criteria. It generally takes into account a negotiation between exerted force and movement. It means small force but large movement. Take a leverage for example, the most common one. A large amount of movement at one-point end with small force, translates into large movement but large force at the other end point.