A heat engine extracts heat (Q_H ) from a thermal reservoir at a temperature of 1000 K and rejects heat (Q_L ) to a thermal reservoir at a temperature of 100 K, while producing work (W). Which one of the combinations of [Q_H , Q_L and W] given is allowed?

The correct option is:
Q_H = 2000 J, Q_L = 750 J, W = 1250 J

Analysis of the Engine Diagram and Thermodynamic Principles:
Based on the provided diagram, the heat engine operates under the following conditions:
1. It extracts heat $Q_H$ from a source at temperature $T_1=1000\text{K}$.
2. It rejects heat $Q_L$ (indicated as $Q_2$ in the image calculations) to a sink at temperature $T_2=100\text{K}$.
3. It performs work output $W$.

For any cyclic process to be possible, it must satisfy two fundamental laws of thermodynamics:

1. First Law of Thermodynamics (Conservation of Energy):
The energy entering the engine must equal the total energy leaving it.

$Q_H=W+Q_L$

2. Second Law of Thermodynamics (Clausius Inequality):
As explicitly shown in the accompanying image, the cyclic integral of heat transfer divided by temperature must be less than or equal to zero for a feasible cycle:

$\oint \frac{dQ}{T}\le 0$
For this two-reservoir system, the inequality becomes:

$\frac{Q_H}{T_1}-\frac{Q_L}{T_2}\le 0$

Evaluating the Given Options:

Option A: Q_H = 2000 J, Q_L = 500 J, W = 1000 J
• First Law check: $W+Q_L=1000+500=1500\text{J}$.
Since $1500\text{J}\ne 2000\text{J}$, this option violates the First Law of Thermodynamics.

Option B (Correct): Q_H = 2000 J, Q_L = 750 J, W = 1250 J
• First Law check: $W+Q_L=1250+750=2000\text{J}$. This satisfies the First Law.
• Second Law check:

$\oint \frac{dQ}{T}=\frac{2000}{1000}-\frac{750}{100}=2-7.5=-5.5\text{J/K}$
Since $-5.5<0$, the Clausius inequality is satisfied, making this a physically possible, irreversible thermodynamic cycle.

Option C: Q_H = 6000 J, Q_L = 500 J, W = 5500 J
• First Law check: $W+Q_L=5500+500=6000\text{J}$. This satisfies the First Law.
• Second Law check:

$\oint \frac{dQ}{T}=\frac{6000}{1000}-\frac{500}{100}=6-5=+1\text{J/K}$
Since $+1>0$, this cycle violates the Second Law of Thermodynamics (it exceeds the maximum possible Carnot efficiency between these temperatures).

Option D: Q_H = 6000 J, Q_L = 600 J, W = 5500 J
• First Law check: $W+Q_L=5500+600=6100\text{J}$.
Since $6100\text{J}\ne 6000\text{J}$, this option violates the First Law of Thermodynamics.