Question

A heat engine operates between a cold reservoir and a hot reservoir. The engine takes 200 J of heat from the hot reservoir and has the efficiency of 0.4. The amount of heat delivered to the cold reservoir in a cycle is:

• A. 100 J
• B. 120 J
• C. 140 J
• D. 160 J
• E. 80 J

Answer 1

The Correct Option is B

Given parameters: \[ Q_H = 200\,\text{J (heat absorbed from hot reservoir)} \] \[ \eta = 0.4 \text{ (efficiency)} \]

Efficiency definition: \[ \eta = \frac{W}{Q_H} \] \[ 0.4 = \frac{W}{200} \] \[ W = 80\,\text{J (work done)} \]

First Law of Thermodynamics: \[ Q_H = W + Q_C \] \[ 200 = 80 + Q_C \]

Heat rejected calculation: \[ Q_C = 200 - 80 \] \[ Q_C = 120\,\text{J} \]

Alternative verification: \[ \eta = 1 - \frac{Q_C}{Q_H} \] \[ 0.4 = 1 - \frac{Q_C}{200} \] \[ \frac{Q_C}{200} = 0.6 \] \[ Q_C = 120\,\text{J} \]

Answer 2

1. Recall the formula for efficiency:

The efficiency (η) of a heat engine is defined as the ratio of the work done (W) by the engine to the heat absorbed (Q_H) from the hot reservoir:

\[\eta = \frac{W}{Q_H}\]

2. Calculate the work done:

We are given \(\eta = 0.4\) and \(Q_H = 200 \, J\). Therefore:

\[W = \eta Q_H = (0.4)(200 \, J) = 80 \, J\]

3. Apply the first law of thermodynamics:

 For a cyclic process, the change in internal energy is zero (ΔU = 0). Therefore:

\[Q = W\]

In a heat engine, the net heat transfer is the heat absorbed from the hot reservoir (Q_H) minus the heat released to the cold reservoir (Q_C):

\[Q = Q_H - Q_C\]

4. Solve for the heat delivered to the cold reservoir:

Since Q = W, we have:

\[W = Q_H - Q_C\]

\[Q_C = Q_H - W = 200 \, J - 80 \, J = 120 \, J\]