Heat and thermodynamics in NET tests temperature scales, calorimetry, the kinetic theory of gases, the first law, and heat engine efficiency. Nearly every numerical here goes wrong for one reason: a temperature left in °C when the formula needs kelvin. Fix that habit and this chapter turns into quick marks.
Key facts
- Physics is 30% of NET-Engineering, a derived 60 of the 200 MCQs.
- Our syllabus data estimates about 6 questions from first-year Heat and Thermodynamics and about 5 from second-year Thermal Physics.
- Syllabus: temperature scales, calorimetry, kinetic theory of gases, and the first law of thermodynamics applied to heat engines.
Source: NUST NET-Engineering composition and the SailToNUST physics syllabus, checked October 2026
What the chapter covers
The first-year chapter runs from temperature and specific heat through the gas laws and kinetic theory to the laws of thermodynamics, heat engines and entropy. The second-year Thermal Physics chapter revisits kinetic theory: the molecular origin of pressure, root-mean-square speed and average kinetic energy. NET mixes both, so a question on v_rms can sit next to one on Carnot efficiency.
Formulas you need
Kelvin T(K) = T(°C) + 273
Fahrenheit F = (9/5)C + 32
A change of 1 °C equals a change of 1 K
Heat Q = mcΔT c in J kg⁻¹ K⁻¹
Latent heat Q = mL L in J kg⁻¹
Mixing (no losses) heat lost by hot = heat gained by cold
Ideal gas PV = nRT R = 8.31 J mol⁻¹ K⁻¹
Fixed amount of gas P₁V₁/T₁ = P₂V₂/T₂
Pressure from KE P = ⅓ρ v_rms²
rms speed v_rms = √(3RT/M) M in kg mol⁻¹
Mean KE per molecule ½m v_rms² = (3/2)kT k = 1.38 × 10⁻²³ J K⁻¹
First law Q = ΔU + W W = work done BY the gas
Work at constant P W = PΔV (in general, the area under the P-V curve)
Isothermal ΔU = 0, so Q = W
Adiabatic Q = 0, so W = −ΔU
Isochoric W = 0, so Q = ΔU
Molar heats Cp − Cv = R
Carnot efficiency η = 1 − T₂/T₁ (kelvin)
Entropy change ΔS = Q/T J K⁻¹
Worked MCQs
1. Doubling volume at constant pressure. An ideal gas at 27 °C is heated at constant pressure until its volume doubles. What is its final temperature?
(a) 54 °C (b) 200 °C (c) 300 °C (d) 327 °C
T₁ = 27 + 273 = 300 K
V ∝ T at constant P, so T₂ = 600 K
T₂ = 600 − 273 = 327 °C
Answer (d). Option (a) is what you get by doubling 27 °C directly.
2. rms speed ratio. The rms speed of a gas is 300 m/s. The molar mass is doubled and the absolute temperature is halved. What is the new rms speed?
(a) 300 m/s (b) 150 m/s (c) 600 m/s (d) 75 m/s
v_rms ∝ √(T/M)
T/M → (½)/(2) = ¼ of the original
v_rms → 300 × √¼ = 150 m/s
Answer (b).
3. Carnot engine. An ideal engine works between 527 °C and 127 °C. What is its efficiency?
(a) 24% (b) 50% (c) 76% (d) 100%
T₁ = 527 + 273 = 800 K, T₂ = 127 + 273 = 400 K
η = 1 − 400/800 = 0.5 = 50%
Answer (b). Using °C gives 1 − 127/527, about 76%, which is option (c). It is there for a reason.
4. First law. A gas absorbs 500 J of heat and does 200 J of work on its surroundings. What is the change in its internal energy?
(a) 700 J (b) 300 J (c) −300 J (d) 200 J
Q = ΔU + W
500 = ΔU + 200
ΔU = 300 J
Answer (b). Check which sign convention the question uses. In the Pakistani textbooks W is the work done by the gas, so heat in and work out pull ΔU in opposite directions.
5. Mixing water. 0.2 kg of water at 80 °C is mixed with 0.3 kg of water at 20 °C. Ignoring losses, what is the final temperature?
(a) 44 °C (b) 50 °C (c) 56 °C (d) 36 °C
0.2 × c × (80 − T) = 0.3 × c × (T − 20)
16 − 0.2T = 0.3T − 6
T = 22 / 0.5 = 44 °C
Answer (a). For the same substance, the final temperature is the mass-weighted average: (0.2 × 80 + 0.3 × 20) / 0.5. The specific heat cancels. Option (b) is the plain average, which only works for equal masses.
Traps that cost marks
- Kelvin everywhere. Gas laws, v_rms, kinetic energy and efficiency all need absolute temperature. Only temperature differences, as in Q = mcΔT, work the same in °C.
- −40 is the same on both scales. −40 °C equals −40 °F. It is a classic one-line MCQ.
- Net work in a cycle is the enclosed area. Clockwise on a P-V diagram means the gas does positive net work. Anticlockwise means work is done on it.
- Isothermal does not mean no heat. In an isothermal expansion the gas absorbs heat and turns all of it into work. ΔU is zero because U of an ideal gas depends on temperature only.
- No engine reaches 100%. That is the second law. A Carnot engine only gets close if the cold reservoir is near 0 K.
- A fan warms a closed room. The motor's electrical energy ends up as heat. You feel cooler because moving air helps sweat evaporate, but the air temperature goes up slightly.
- Entropy rises when ice melts. Heat flows in at constant temperature, so ΔS = mL/T is positive.
Speed tips for the exam
Convert to kelvin in the margin before you read the options. Write "300 K" beside "27 °C" the moment you see it.
For ratio questions on gases, cross out what is held fixed and write the proportionality: V ∝ T at fixed P, P ∝ T at fixed V, v_rms ∝ √(T/M). Then the arithmetic is one step.
For P-V diagrams, count grid squares for the enclosed area instead of splitting the shape into triangles and rectangles.
Kinetic theory overlaps with chemistry's gas laws, so revise them together. For the full physics plan, see the NET physics guide, and for every formula on one page, the physics formula sheet.
FAQ
Which sign convention does NET use for the first law?
FSc textbooks in Pakistan write Q = ΔU + W, with W as work done by the gas. Use that convention unless a question defines W as work done on the gas, in which case follow the question.
What is the difference between Cp and Cv?
Cp is the molar heat capacity at constant pressure and Cv at constant volume. Cp is larger because at constant pressure some heat goes into expansion work. For an ideal gas, Cp − Cv = R.
Does the rms speed depend on pressure?
Not directly. v_rms = √(3RT/M) depends on temperature and molar mass only. Changing pressure at constant temperature leaves it unchanged.
Is thermodynamics hard in NET?
The ideas are short and the numbers are easy. Students lose marks to units and sign conventions far more than to difficult physics.
