🔥 Thermodynamics Basics — Heat Transfer & Gas Laws
Thermodynamics governs every heat engine, refrigerator, boiler and HVAC system. The specific heat capacity equation Q = mcΔT and the ideal gas laws let engineers size heat exchangers, calculate energy budgets and predict how gases behave under changing conditions.
2Topics
8Examples
4Calculators
12Practice Qs
🔥 Level 3 context: Temperature for gas law calculations must always be in Kelvin: K = °C + 273. Using Celsius in gas law ratios gives completely wrong answers — this is one of the most common errors in thermodynamics problems.
🔥Heat & Specific HeatQ = mcΔT, latent heat
🌸Gas LawsBoyle, Charles, combined
⚙️Applications4 thermodynamics calculators
🎮Game12 randomised questions
📝Practice12 graded questions
🔥
Heat Transfer & Specific Heat Capacity
Heat energy transferred to or from a substance depends on its mass, its specific heat capacity (a material property) and the temperature change. Latent heat covers phase changes where temperature stays constant.
🎮Live explorer — Q = mcΔTDrag sliders — heat energy updates
Heat Q
167.4 kJ
Q = mcΔT
Time at 1 kW
167.4 s
t = Q / P
Power for 60 s
2790 W
P = Q / t
💡 A temperature change of 1 K equals a change of 1°C — so ΔT is the same whether in Kelvin or Celsius.
📋Try it — Heat & Temperature CalculatorUpdates as you type
Select what to find and enter values above.
📚 Key vocabulary
- Specific heat capacity (c)
- energy needed to raise 1 kg of a substance by 1 K. Unit: J/kg·K. A material constant.
- Sensible heat
- heat that causes a measurable temperature change — calculated using Q = mcΔT.
- Latent heat (L)
- heat energy absorbed or released during a phase change at constant temperature. Q = mL.
- Latent heat of fusion (L𝑓)
- energy to melt (or freeze) 1 kg of a substance. Water: L𝑓 = 334 kJ/kg.
- Latent heat of vaporisation (L𝑣)
- energy to boil (or condense) 1 kg. Water: L𝑣 = 2260 kJ/kg.
- Thermal power (P)
- rate of heat transfer. P = Q/t (watts). Used to size heaters, coolers and heat exchangers.
👁️ Visual aid — heating curve for water: sensible heat raises temperature; latent heat drives phase changes at 0°C and 100°C
✍️Worked example 1 — heating water step through it
QuestionCalculate the heat needed to raise 3 kg of water from 20°C to 80°C. (c𝑣𝑎𝑡𝑒𝑟 = 4186 J/kg·K)
Formula
Q = mcΔTValues
m = 3 kg c = 4186 J/kg·K ΔT = 80 − 20 = 60 KSubstitute
Q = 3 × 4186 × 60Calculate
= 12 558 × 60 = 753 480 J = 753.5 kJAnswerQ = 753.5 kJ ✓
Step 1 of 6
✍️Worked example 2 — find mass from heat energy
QuestionAn electric heater supplies 90 kJ to raise an aluminium block by 100°C. (c = 900 J/kg·K) Find the mass.
Rearrange
m = Q ÷ (c × ΔT)Substitute
m = 90 000 ÷ (900 × 100) = 90 000 ÷ 90 000Answerm = 1.0 kg
✍️Worked example 3 — latent heat of fusion
QuestionCalculate the heat needed to melt 2.5 kg of ice at 0°C. (L𝑓 = 334 000 J/kg)
Formula
Q = mLSubstitute
Q = 2.5 × 334 000 = 835 000 JAnswerQ = 835 kJ
✍️Worked example 4 — heater power and time
QuestionA 2 kW heater warms 5 kg of steel (c = 500 J/kg·K) from 20°C to 220°C. Find the heat required and time taken.
Heat Q
Q = 5 × 500 × 200 = 500 000 J = 500 kJTime
t = Q ÷ P = 500 000 ÷ 2000 = 250 sAnswerQ = 500 kJ t = 250 s
🌸
Gas Laws — Boyle, Charles & Combined
The ideal gas laws describe how pressure, volume and temperature of a fixed mass of gas are related. They are essential for pneumatics, hydraulics, compressed air systems and thermodynamic cycle analysis.
🎮Live explorer — Boyle's LawDrag sliders — volume or pressure updates
P₁V₁
1000 kPa·L
constant (isothermal)
New volume V₂
5.0 L
V₂ = P₁V₁/P₂
Ratio V₂/V₁
0.50
compression ratio
💡 Boyle's Law: pressure and volume are inversely proportional at constant temperature. Double the pressure → half the volume.
📋Try it — Gas Laws CalculatorUpdates as you type
Select a gas law and enter the known values.
📚 Key vocabulary
- Absolute temperature (K)
- temperature measured from absolute zero. K = °C + 273. Must be used in all gas law calculations.
- Boyle's Law
- at constant temperature, P × V = constant. A gas compressed to half its volume doubles in pressure.
- Charles's Law
- at constant pressure, V/T = constant. Volume is directly proportional to absolute temperature.
- Gay-Lussac's Law
- at constant volume, P/T = constant. Pressure is directly proportional to absolute temperature.
- Isothermal process
- a process at constant temperature. Boyle's Law applies.
- Isobaric process
- a process at constant pressure. Charles's Law applies. Common in piston-cylinder devices open to atmosphere.
👁️ Visual aid — the three gas laws compared. In each case one variable is held constant while the other two change.
✍️Worked example 5 — Boyle's Law step through it
QuestionA gas occupies 8 litres at 150 kPa. The pressure increases to 400 kPa at constant temperature. Find the new volume.
Law
Boyle's Law: P₁V₁ = P₂V₂ (temperature constant)Rearrange
V₂ = P₁V₁ ÷ P₂Substitute
V₂ = (150 × 8) ÷ 400 = 1200 ÷ 400AnswerV₂ = 3 litres ✓
Step 1 of 5
✍️Worked example 6 — Charles's Law
QuestionA gas has volume 6 litres at 27°C. It is heated to 127°C at constant pressure. Find the new volume.
Convert
T₁ = 27 + 273 = 300 K T₂ = 127 + 273 = 400 KCharles
V₂ = V₁ × T₂ ÷ T₁ = 6 × 400 ÷ 300AnswerV₂ = 8 litres
✍️Worked example 7 — Gay-Lussac's Law
QuestionA sealed container holds gas at 200 kPa and 17°C. It is heated to 77°C. Find the new pressure.
Convert
T₁ = 290 K T₂ = 350 KGay-Lussac
P₂ = P₁ × T₂ ÷ T₁ = 200 × 350 ÷ 290 = 241.4 kPaAnswerP₂ = 241.4 kPa
✍️Worked example 8 — Combined Gas Law
QuestionA gas at 100 kPa, 5 litres, 27°C is compressed to 250 kPa at 77°C. Find the new volume.
Convert
T₁ = 300 K T₂ = 350 KCombined
V₂ = P₁V₁T₂ ÷ (T₁P₂) = (100 × 5 × 350) ÷ (300 × 250)Calculate
= 175 000 ÷ 75 000 = 2.333AnswerV₂ = 2.33 litres
⚙️
Engineering Applications
Four thermodynamics calculators for practical heating, cooling and gas system scenarios.
🔥 Water heater sizing
Find the heat energy required and heater power needed to raise water temperature in a given time.
Q = mcΔT P = Q ÷ t
🧮 Calculator
Enter mass, temperature rise and time.
🌸 Boyle's Law — compressed air
Find the new volume when a compressed air cylinder is opened to a lower pressure.
V₂ = P₁V₁ ÷ P₂
🧮 Calculator
Enter P₁, V₁ and P₂.
🔥 Latent heat — phase change
Calculate the heat energy released or absorbed when a substance changes phase.
Q = mL
🧮 Calculator
Enter mass and latent heat value.
🌸 Combined Gas Law
Find the final volume when pressure and temperature both change for a fixed mass of ideal gas.
V₂ = P₁V₁T₂ ÷ (P₂T₁)
🧮 Calculator
Enter P₁, V₁, T₁, P₂ and T₂.
🎮
Thermodynamics Challenge
12 randomised questions — heat, specific heat capacity and gas laws.
🔥 Quick-fire round
Question 1 / 12
Score: 0
Streak: 0 🔥
Press Start to begin!
📝
Practice Questions
12 graded questions — type your answer to check, then reveal full working.
🔥 State the formula used, convert temperatures to Kelvin where needed, and show every step. c𝑣𝑎𝑡𝑒𝑟 = 4186 J/kg·K c𝑠𝑡𝑒𝑒𝑙 = 500 J/kg·K L𝑓𝑢𝑠𝑖𝑜𝑛 = 334 000 J/kg.
Q1 · Q = mcΔT1 mark
Calculate the heat needed to raise 4 kg of water from 15°C to 65°C. (c = 4186 J/kg·K)
Q = mcΔT = 4 × 4186 × 50 = 837 200 J = 837.2 kJ
Q2 · Q = mcΔT2 marks
A 2.5 kg steel shaft (c = 500 J/kg·K) is heated from 20°C to 320°C. Find the heat absorbed.
Q = 2.5 × 500 × 300 = 375 000 J = 375 kJ
Q3 · Find mass2 marks
126 kJ heats aluminium (c = 900 J/kg·K) from 20°C to 160°C. Find the mass.
m = Q/(cΔT) = 126 000/(900×140) = 126 000/126 000 = 1.0 kg
Q4 · Find ΔT2 marks
50 kJ is supplied to 2 kg of copper (c = 390 J/kg·K). Find the temperature rise.
ΔT = Q/(mc) = 50 000/(2×390) = 50 000/780 = 64.1°C
Q5 · Latent heat2 marks
Calculate the heat released when 3 kg of water freezes. (L𝑓 = 334 000 J/kg)
Q = mL = 3 × 334 000 = 1 002 000 J = 1002 kJ
Q6 · Heater power2 marks
A 3 kW heater heats 4 kg of water (ΔT = 60°C, c = 4186 J/kg·K). How many seconds does it take?
Q = 4×4186×60 = 1 004 640 J
t = Q/P = 1 004 640/3000 = 334.9 s
t = Q/P = 1 004 640/3000 = 334.9 s
Q7 · Boyle's Law2 marks
A gas has volume 12 litres at 100 kPa. The pressure rises to 300 kPa at constant temperature. Find the new volume.
V₂ = P₁V₁/P₂ = 100×12/300 = 4 litres
Q8 · Charles's Law2 marks
A gas occupies 4 litres at 27°C. It is heated to 177°C at constant pressure. Find the new volume.
T₁=300K T₂=450K
V₂ = V₁×T₂/T₁ = 4×450/300 = 6 litres
V₂ = V₁×T₂/T₁ = 4×450/300 = 6 litres
Q9 · Gay-Lussac's Law2 marks
A sealed cylinder holds gas at 150 kPa and 27°C. It is heated to 127°C. Find the new pressure.
T₁=300K T₂=400K
P₂ = P₁×T₂/T₁ = 150×400/300 = 200 kPa
P₂ = P₁×T₂/T₁ = 150×400/300 = 200 kPa
Q10 · Combined gas law3 marks
A gas at 200 kPa, 3 litres, 27°C is heated to 127°C and the pressure rises to 300 kPa. Find the new volume.
T₁=300K T₂=400K
V₂ = P₁V₁T₂/(P₂T₁) = 200×3×400/(300×300) = 240000/90000 = 2.667 litres
V₂ = P₁V₁T₂/(P₂T₁) = 200×3×400/(300×300) = 240000/90000 = 2.667 litres
Q11 · Combined3 marks
An aluminium casting (m = 3 kg, c = 900 J/kg·K) cools from 650°C to 50°C in a mould. Find (a) heat released, (b) average cooling power if it takes 300 s.
Q = 3×900×600 = 1 620 000 J = 1620 kJ
P = Q/t = 1 620 000/300 = 5400 W = 5.4 kW
P = Q/t = 1 620 000/300 = 5400 W = 5.4 kW
Q12 · Combined3 marks
A gas at 80 kPa, 10 litres, 17°C is compressed to 4 litres at 57°C. Find the final pressure.
T₁ = 17+273 = 290 K T₂ = 57+273 = 330 K
P₂ = P₁V₁T₂ ÷ (V₂T₁) = 80×10×330 ÷ (4×290) = 264 000 ÷ 1160 = 227.6 kPa
P₂ = P₁V₁T₂ ÷ (V₂T₁) = 80×10×330 ÷ (4×290) = 264 000 ÷ 1160 = 227.6 kPa