Heat Engine Numerical Problems Solved – Step-by-Step for RRB ALP Aspirants

Heat Engine numericals in RRB ALP CBT-2 are extremely predictable. Most questions revolve around:

• Efficiency formula

• Heat supplied & heat rejected

• Work output

• Carnot engine efficiency

• Temperature conversion (°C → K)

This guide gives you step-by-step solved problems, formula explanations, and common RRB exam tricks so you can solve questions in 5–10 seconds during the exam.

Important Formulas (Must Memorize)

Keep this table handy while solving the numericals below.

Solved Numerical Problems (RRB ALP PYQ Style)

Problem 1: Find Efficiency from Heat Supplied & Heat Rejected

A heat engine receives 1000 J heat and rejects 400 J. Calculate efficiency.

Step-by-Step Solution

Given:
Q₁ = 1000 J
Q₂ = 400 J

Formula:

$$\eta = 1 - \frac{Q_2}{Q_1}$$ $$\eta = 1 - \frac{400}{1000}$$ $$\eta = 1 - 0.4 = 0.6 = 60\%$$

Final Answer: ✔ 60%

Problem 2: Calculate Work Output

If a heat engine receives 1500 J heat and its efficiency is 40%, find the work output.

Step-by-Step Solution

Given:
Q₁ = 1500 J
η = 40% = 0.40

Formula:

$$W = \eta \times Q_1$$ $$W = 0.40 \times 1500 = 600\ J$$

Final Answer: ✔ 600 J

Problem 3: Find Heat Rejected (Q₂)

A heat engine has an efficiency of 35%. If heat supplied is 1200 J, find heat rejected.

Step-by-Step Solution

Given:
η = 35% = 0.35
Q₁ = 1200 J

Formula (work output):

$$W = \eta Q_1 = 0.35 \times 1200 = 420 J$$

Heat rejected:

$$Q_2 = Q_1 - W$$ $$Q_2 = 1200 - 420 = 780 J$$

Final Answer: ✔ 780 J

Problem 4: Carnot Efficiency (Temperature-based)

A Carnot engine works between 600 K and 300 K. Find efficiency.

Step-by-Step Solution

Given:
T₁ = 600 K
T₂ = 300 K

Formula:

$$\eta = 1 - \frac{T_2}{T_1}$$ $$\eta = 1 - \frac{300}{600}$$ $$\eta = 1 - 0.5 = 0.5 = 50\%$$

Final Answer: ✔ 50%

Problem 5: Temperature Conversion Trap (RRB Favorite)

A Carnot engine operates between 327°C and 27°C. Find efficiency.

Step-by-Step Solution

Convert to Kelvin:

$$T_1 = 327 + 273 = 600\ K$$ $$T_2 = 27 + 273 = 300\ K$$

Now apply the formula:

$$\eta = 1 - \frac{300}{600} = 0.5 = 50\%$$

Final Answer: ✔ 50%

Problem 6: Efficiency When Work Output is Given

A heat engine produces 500 J of work using 2000 J of heat. Find its efficiency.

Step-by-Step Solution

Given:
W = 500 J
Q₁ = 2000 J

Formula:

$$\eta = \frac{W}{Q_1}$$ $$\eta = \frac{500}{2000} = 0.25 = 25\%$$

Final Answer: ✔ 25%

Problem 7: Find Heat Supplied When Efficiency & Heat Rejected Are Known

Efficiency = 30%, Heat rejected = 700 J. Find heat supplied.

Step-by-Step Solution

Given:
η = 0.30
Q₂ = 700 J

Formula:

$$\eta = 1 - \frac{Q_2}{Q_1}$$

Rearrange to calculate Q₁:

$$1 - \eta = \frac{Q_2}{Q_1}$$ $$Q_1 = \frac{Q_2}{1 - \eta}$$ $$Q_1 = \frac{700}{1 - 0.30}$$ $$Q_1 = \frac{700}{0.70} = 1000\ J$$

Final Answer: ✔ 1000 J

Problem 8: Find Efficiency When Work & Heat Rejected Are Given

A heat engine does 200 J of work and rejects 300 J heat. Find efficiency.

Step-by-Step Solution

Work done:
W = 200 J
Heat rejected:
Q₂ = 300 J

Find heat supplied:

$$Q_1 = W + Q_2$$ $$Q_1 = 200 + 300 = 500 J$$

Now efficiency:

$$\eta = \frac{200}{500} = 0.4 = 40\%$$

Final Answer: ✔ 40%

Quick Revision Table (Exam Day Sheet)

RRB ALP Tips to Solve Numericals Faster

✔ 1. Always convert °C → K

RRB intentionally twists this.

✔ 2. Use fractions (1/2, 1/3) to speed up

Avoid decimals.

✔ 3. Memorize 5–6 common ratios

RRB repeats values like 1000–400, 600–300, 800–200.

✔ 4. Avoid silly mistakes

Most students lose marks due to Q₁–Q₂ confusion.

✔ 5. Practice 20 numericals daily

Heat Engine becomes super easy with repetition.

Exercise Questions (Try Yourself)

1. Q₁ = 900 J, Q₂ = 300 J. Find η.

2. T₁ = 500 K, T₂ = 200 K. Find efficiency.

3. W = 250 J, Q₁ = 800 J. Find efficiency.

4. Q₁ = 600 J, η = 40%. Find Q₂.

5. T₁ = 227°C, T₂ = 27°C. Find efficiency (Kelvin required).

Final thoughts

Heat Engine numericals are one of the easiest scoring areas in RRB ALP CBT-2. With the formulas, solved examples, and tricks provided here, you can easily solve most questions in 5–10 seconds during the exam.