Class 12 - Unit # 17 : Second Law of Thermodynamics -Solved Numericals

 

Problem 1

A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. How much heat is discarded, how much work does the engine do, and what is the efficiency? (Ans: 1400 J, 600 J, 30%)

Given Data:

• Heat Absorbed (Q₁) = 2000 J
• Temperature of Hot Reservoir (T₁) = 500 K
• Temperature of Cold Reservoir (T₂) = 350 K
• Heat Discarded (Q₂) = ?
• Work Done (W) = ?
• Efficiency (η) = ?

Solution:

The efficiency of a Carnot engine in terms of absolute temperatures is given by:

η = (1 -

T₂

T₁

) × 100%

Substituting values:

η = (1 -

350

500

) × 100%

η = (1 - 0.7) × 100% = 0.3 × 100%

η = 30%

For a Carnot engine, the heat ratio equals the temperature ratio:

Q₂

Q₁

=

T₂

T₁

Q₂ = Q₁ ×

T₂

T₁

Q₂ = 2000 ×

350

500

= 2000 × 0.7

Q₂ = 1400 J

Now, calculate Work Done (W):

W = Q₁ - Q₂

W = 2000 J - 1400 J

W = 600 J

Result:

The Efficiency of the engine is 30%, heat rejected is 1400 J, and work done by the engine is 600 J.

Problem 2

One kilogram of ice at 0°C is melted and converted to water at 0°C. Compute its change in entropy. (Ans: 1230.76 J·K^-1)

Given Data:

• Mass of Ice (m) = 1 kg
• Temperature (T) = 0°C + 273 = 273 K
• Latent Heat of Fusion of Ice (H_f) = 336,000 J/kg
• Change in Entropy (ΔS) = ?

Solution:

First, find the total heat required (ΔQ) to melt the ice:

ΔQ = m × H_f

ΔQ = 1 kg × 336,000 J/kg = 336,000 J

Now, find the change in entropy (ΔS):

ΔS =

ΔQ

T

ΔS =

336,000

273

ΔS = 1230.77 J/K

Result:

The change in entropy is 1230.77 J·K^-1.

Problem 3

In a high-pressure steam turbine engine, the steam is heated to 600°C and exhausted at about 90°C. What is the highest possible efficiency of any engine that operates between these two temperatures? (Ans: 58.4%)

Given Data:

• Temperature of Hot Source (T₁) = 600°C + 273 = 873 K
• Temperature of Sink (T₂) = 90°C + 273 = 363 K
• Maximum Efficiency (η_max) = ?

Solution:

The maximum theoretical efficiency is given by the Carnot efficiency formula:

η =

T₁ - T₂

T₁

× 100%

Substituting the thermodynamic temperatures:

η =

873 - 363

873

× 100%

η =

510

873

× 100% = 0.5841 × 100%

η = 58.41%

Result:

The highest efficiency of this engine is 58.41%.

Problem 4

Temperature difference between the surface water and bottom water in Manchester Lake might be 5°C. Assuming the surface water to be at 20°C, what highest efficiency a steam engine could have if it operates between these two temperatures? (Ans: 1.71%)

Given Data:

• Temperature of Hot Body (T₁) = 20°C + 273 = 293 K
• Temperature Difference (ΔT) = 5°C = 5 K
• Temperature of Cold Body (T₂) = T₁ - ΔT = 20°C - 5°C = 15°C + 273 = 288 K
• Maximum Efficiency (η) = ?

Solution:

Using the engine efficiency formula:

η =

T₁ - T₂

T₁

× 100%

η =

5

293

× 100% = 0.01706 × 100%

η = 1.71%

Result:

The highest efficiency of this engine is 1.71%.

Problem 5

A heat engine works at the rate of 500 kW. The efficiency of the engine is 30%. Calculate the loss of heat per hour. (Ans: 4.2 × 10⁹ J)

Given Data:

• Output Power (P) = 500 kW = 5 × 10⁵ W
• Efficiency (η) = 30% = 0.30
• Time (t) = 1 hour = 3600 seconds
• Loss of Heat (Q₂) = ?

Solution:

First, find the output work done (W) in one hour:

W = P × t

W = (5 × 10⁵ W) × 3600 s = 1.8 × 10⁹ J

Now, use the efficiency definition to calculate total heat input (Q₁):

η =

W

Q₁

⇒ Q₁ =

W

η

Q₁ =

1.8 × 10⁹

0.30

= 6 × 10⁹ J

Finally, calculate the heat rejected/lost (Q₂):

Q₂ = Q₁ - W

Q₂ = (6 × 10⁹) - (1.8 × 10⁹)

Q₂ = 4.2 × 10⁹ J

Result:

The heat lost per hour is equal to 4.2 × 10⁹ J.

Problem 6

A heat engine performs work at a rate of 0.4166 Watts for one hour and rejects 4500 J of heat to the sink. What is the efficiency of the engine? (Ans: 25%)

Given Data:

• Power (P) = 0.4166 W
• Time (t) = 1 hour = 3600 s
• Heat Rejected (Q₂) = 4500 J
• Efficiency (η) = ?

Solution:

Find the work done (W) by the engine during this time:

W = P × t = 0.4166 × 3600 ≈ 1500 J

Calculate total heat absorbed from the source (Q₁):

Q₁ = W + Q₂ = 1500 + 4500 = 6000 J

Now, calculate the efficiency (η):

η =

W

Q₁

× 100%

η =

1500

6000

× 100% = 0.25 × 100%

η = 25%

Result:

The efficiency of the engine is 25%.

Problem 7

A Carnot engine operates between temperatures of 850 K and 300 K. The engine performs 1200 J of work in each cycle, which takes 0.25 s. Calculate:
(a) Efficiency of the engine
(b) Average power output
(c) Heat energy extracted from the high-temperature reservoir
(d) Heat energy rejected to the low-temperature reservoir
[Ans: (a) 64.7% (b) 4.8 kW (c) 1854.7 J (d) 654.7 J]

Given Data:

• Hot Reservoir Temperature (T₁) = 850 K
• Cold Reservoir Temperature (T₂) = 300 K
• Work Done per Cycle (W) = 1200 J
• Time Period of Cycle (t) = 0.25 s

Solution:

(a) Efficiency (η)

η = (1 -

T₂

T₁

) × 100%

η = (1 -

300

850

) × 100% = (1 - 0.3529) × 100%

η = 64.71%

(b) Average Power (P)

P =

W

t

=

1200

0.25

P = 4800 W = 4.8 kW

(c) Heat Energy Extracted (Q₁)

η =

W

Q₁

⇒ Q₁ =

W

η

=

1200

0.6471

Q₁ = 1854.43 J

(d) Heat Delivered to Sink (Q₂)

Q₂ = Q₁ - W = 1854.43 - 1200

Q₂ = 654.43 J

Result:

(a) Engine efficiency is 64.71%.
(b) Average power output is 4.8 kW.
(c) Heat taken from high temperature source is 1854.43 J.
(d) Heat rejected to low temperature reservoir is 654.43 J.

Problem 8

A Carnot engine absorbs 52 kJ as heat and exhausts 36 kJ as heat in each cycle. Calculate: (a) The engine efficiency (b) The work done per cycle in kilojoules. [Ans: (a) 30.76% (b) 16 kJ]

Given Data:

• Heat Absorbed (Q₁) = 52 kJ
• Heat Discarded (Q₂) = 36 kJ
• Efficiency (η) = ?
• Work Done (W) = ?

Solution:

(a) Engine Efficiency (η)

η =

Q₁ - Q₂

Q₁

× 100%

η =

52 - 36

52

× 100% =

16

52

× 100%

η = 30.76%

(b) Work Done per Cycle (W)

W = Q₁ - Q₂

W = 52 kJ - 36 kJ

W = 16 kJ

Result:

The engine efficiency is 30.76% and the work done per cycle is 16 kJ.