Question

A carnot engine takes in 3000 K Cal of heat from reservoir at 627°C and gives it to a sink at 25°C. The work done by the engine is -   
1. 4.2 × 10⁶ Joule
2. 8.4 × 10⁶ Joule
3. 16.8 × 10⁶ Joule
4. Zero

Answer 1

Correct Answer - Option 2 : 8.4 × 10⁶ Joule

CONCEPT:

The efficiency of the Carnot cycle (η):

• It is defined as the ratio of net mechanical work done per cycle the gas (W) to the amount of heat energy absorbed per cycle from the source (Q1) i.e.,

\(η = \frac{W}{{{Q_1}}}\;\)

As work is done by the engine per cycle  is

W = Q1 – Q2

Where, Q1 = amount of heat energy absorbed per cycle from the source and Q2 = energy absorbed per cycle from the sink.

\( ⇒ η = \frac{{{Q_1} - {Q_2}}}{{{Q_1}}} = 1 - \frac{{{Q_2}}}{{{Q_1}}}\)

As, \(\frac{{{Q_2}}}{{{Q_1}}} = \frac{{{T_2}}}{{{T_1}}}\)

\( ⇒ η = 1 - \frac{{{T_2}}}{{{T_1}}}\)

Where T1 = temperature of the sink and T2 = temperature of the source.

CALCULATION:

Given - 

Q1 = 3000 kcal = 3 × 106 cal

T1 = 627 + 273 = 900 K

T2 = 25 + 273 = 298 K;

Now the efficiency will be 

\(η = 1- \frac {298}{900} = 0.668\)

Now the work done will  be 

W = η × Q₁;

⇒ W = 0.668 × 3 × 10⁶ cal

⇒ W = 2 × 10⁶ cal

For converting it to joules, 1 kcal = 4.18 kJ;

⇒ W = 2006.67 × 4.18 kJ;

⇒ W = 8387.867 kJ

⇒ W ≈ 8.4 × 10⁶ kJ;