Thermodynamics Solved Examples (11 Physics)
Thermodynamics studies heat, work, and internal energy. These examples cover the first law, heat capacity, specific heat, and applications involving temper
TL;DR: Thermodynamics studies heat, work, and internal energy. These examples cover the first law, heat capacity, specific heat, and applications involving t…
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Thermodynamics studies heat, work, and internal energy. These examples cover the first law, heat capacity, specific heat, and applications involving temper
Thermodynamics — Solved Numerical Examples (Step by Step)
Example 1: Calculate the heat required to raise the temperature of 2 kg of water from 20°C to 50°C. (Specific heat of water = 4200 J/kg°C)
Solution: Use the formula: Q = m × c × ΔT
Where Q = heat energy, m = mass, c = specific heat, ΔT = change in temperature
Given: m = 2 kg, c = 4200 J/kg°C, initial temperature = 20°C, final temperature = 50°C
ΔT = 50 - 20 = 30°C
Q = 2 × 4200 × 30
Q = 252,000 J = 252 kJ
Example 2: A metal block of mass 0.5 kg is heated from 25°C to 75°C, absorbing 10,000 J of heat. Find the specific heat capacity of the metal.
Solution: From Q = m × c × ΔT, we get:
c = Q / (m × ΔT)
Given: Q = 10,000 J, m = 0.5 kg, initial T = 25°C, final T = 75°C
ΔT = 75 - 25 = 50°C
c = 10,000 / (0.5 × 50)
c = 10,000 / 25
c = 400 J/kg°C
Example 3: An ideal gas undergoes an isothermal expansion at 300 K. If initial volume is 1 m³ and final volume is 2 m³, find the work done by the gas. (Pressure remains constant at 1 atm = 101,325 Pa)
Solution: For isothermal process: PV = constant
For an isothermal expansion, work done:
W = nRT ln(V_f / V_i) = P_i V_i ln(V_f / V_i)
Given: V_i = 1 m³, V_f = 2 m³, P = 101,325 Pa
W = 101,325 × 1 × ln(2 / 1)
W = 101,325 × ln(2)
W = 101,325 × 0.693
W ≈ 70,218 J ≈ 70.2 kJ
Example 4: Calculate the heat released when 100 g of steam at 100°C condenses to water at 100°C. (Latent heat of vaporization = 2,260 kJ/kg)
Solution: For phase change (condensation):
Q = m × L
Where L is latent heat of vaporization
Given: m = 100 g = 0.1 kg, L = 2,260 kJ/kg
Q = 0.1 × 2,260
Q = 226 kJ
Negative sign indicates heat is released.
Heat released = 226 kJ
Example 5: A system absorbs 500 J of heat and does 200 J of work on surroundings. Find the change in internal energy.
Solution: First law of thermodynamics:
ΔU = Q - W
Where ΔU = change in internal energy, Q = heat absorbed, W = work done by system
Given: Q = 500 J (heat absorbed), W = 200 J (work done by system)
ΔU = 500 - 200 = 300 J
The internal energy of the system increases by 300 J.
Example 6: Find the final temperature when 500 g of ice at 0°C is mixed with 1000 g of water at 80°C. (Latent heat of fusion = 334 kJ/kg, specific heat of water = 4.2 kJ/kg°C)
Solution: Heat released by warm water = Heat absorbed by ice (fusion) + Heat absorbed by melted ice
Let final temperature be T°C.
Heat released by 1000 g water cooling from 80°C to T°C:
Q1 = 1 × 4.2 × (80 - T)
Heat absorbed by 500 g ice melting:
Q2 = 0.5 × 334 = 167 kJ
Heat absorbed by melted ice warming from 0°C to T°C:
Q3 = 0.5 × 4.2 × T
Energy balance: Q1 = Q2 + Q3
1 × 4.2 × (80 - T) = 167 + 0.5 × 4.2 × T
4.2(80 - T) = 167 + 2.1T
336 - 4.2T = 167 + 2.1T
169 = 6.3T
T = 26.8°C ≈ 27°C
Example 7: A gas with volume 2 m³ at pressure 100 kPa and temperature 300 K is compressed adiabatically. Find the work done on the gas if the final temperature is 400 K. (Assume Cv = 5R/2)
Solution: For adiabatic process: Q = 0
First law: ΔU = -W (work done on gas is negative of work done by gas)
For ideal gas: ΔU = n × Cv × ΔT
First find number of moles:
PV = nRT
n = PV / RT = (100,000 × 2) / (8.314 × 300)
n = 200,000 / 2,494.2
n ≈ 80.2 moles
ΔU = 80.2 × (5/2 × 8.314) × (400 - 300)
ΔU = 80.2 × 20.785 × 100
ΔU ≈ 166,700 J ≈ 166.7 kJ
Work done on gas = 166.7 kJ
Tips
- Q = m × c × ΔT is used when temperature changes without phase change.
- Q = m × L is used for phase changes (melting, vaporization, condensation, freezing).
- In calorimetry problems, heat lost by one object equals heat gained by another (conservation of energy).
- First law: ΔU = Q - W; W positive when gas does work, negative when work is done on gas.
Frequently Asked Questions
What is the difference between heat and temperature?
Temperature measures the average kinetic energy of molecules. Heat is the transfer of thermal energy between objects. A large object at low temperature can contain more heat than a small object at high temperature.
Why does latent heat involve no temperature change?
During phase change (melting, vaporization), the absorbed heat breaks intermolecular bonds rather than increasing kinetic energy. Temperature remains constant because the energy goes into changing the phase, not motion of molecules.
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