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 answer is: Q_H = 2000 J, Q_L = 750 J, W = 1250 J

Step-by-Step Explanation:

To determine which combination of heat input (Q_H), heat rejection (Q_L), and work output (W) is allowed for the heat engine, we must satisfy both the First and Second Laws of Thermodynamics.

As illustrated in the provided schematic diagram, the heat engine acts as a cyclic system operating between a high-temperature source reservoir at
T₁ = T_H = 1000 K
and a low-temperature sink reservoir at
T₂ = T_L = 100 K.

1. First Law of Thermodynamics (Energy Conservation):
For a cyclic process, the net work output must equal the net heat added to the system:

$W=Q_H-Q_L$

which can also be written as:

$Q_H=W+Q_L$

2. Second Law of Thermodynamics (Clausius Inequality):
For any thermodynamic cycle to be physically feasible, the cyclic integral of heat transfer divided by temperature must be less than or equal to zero:

$\oint \frac{dQ}{T}\le 0$

For our two-reservoir heat engine, heat Q_H is absorbed from the source at T_H (positive heat transfer) and heat Q_L is rejected to the sink at T_L (negative heat transfer). Thus, the inequality becomes:

$\frac{Q_H}{T_H}-\frac{Q_L}{T_L}\le 0$

Substituting the given reservoir temperatures T_H = 1000 K and T_L = 100 K, the feasibility condition is:

$\frac{Q_H}{1000}-\frac{Q_L}{100}\le 0$

Let's evaluate each option using these two rules:

Option 1: Q_H = 2000 J, Q_L = 500 J, W = 1000 J
First Law check:

$W+Q_L=1000\text{ J}+500\text{ J}=1500\text{ J}\ne Q_H\text{ (2000 J)}$

Since this violates the conservation of energy (First Law), this combination is not allowed.

Option 2: Q_H = 2000 J, Q_L = 750 J, W = 1250 J
First Law check:

$W+Q_L=1250\text{ J}+750\text{ J}=2000\text{ J}=Q_H$

The First Law is satisfied.

Second Law check (Clausius Inequality):

$\oint \frac{dQ}{T}=\frac{2000}{1000}-\frac{750}{100}=2-7.5=-5.5\text{ J/K}$

Since
-5.5 J/K < 0
, the Clausius inequality is satisfied (indicating an irreversible but feasible cycle). Therefore, this combination is allowed.

Option 3: Q_H = 6000 J, Q_L = 500 J, W = 5500 J
First Law check:

$W+Q_L=5500\text{ J}+500\text{ J}=6000\text{ J}=Q_H$

The First Law is satisfied.

Second Law check (Clausius Inequality):

$\oint \frac{dQ}{T}=\frac{6000}{1000}-\frac{500}{100}=6-5=+1\text{ J/K}$

Since
+1 J/K > 0
, this violates the Clausius Inequality (impossible cycle). Therefore, this combination is not allowed.

Option 4: Q_H = 6000 J, Q_L = 600 J, W = 5500 J
First Law check:

$W+Q_L=5500\text{ J}+600\text{ J}=6100\text{ J}\ne Q_H\text{ (6000 J)}$

Since this violates the conservation of energy (First Law), this combination is not allowed.