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  3. Zeroth Law of Thermodynamics

Thermodynamics · JEE & NEET Physics

Zeroth Law of Thermodynamics: notes and previous year questions

Thermal equilibrium, the zeroth law, temperature versus heat, and the Celsius, Fahrenheit and Kelvin scales.

Zeroth Law of Thermodynamics in short

  • Bodies in thermal equilibrium have the same temperature and exchange no net heat.
  • By itself, heat flows only from a hotter body to a colder one.
  • Zeroth law: if A and B are each in thermal equilibrium with C, they are in equilibrium with each other.
  • The zeroth law defines temperature and makes thermometers possible (the thermometer is C).

1Hot meets cold

Put a hot block (80 °C) against a cold one (20 °C). Energy flows from the hot block into the cold one: the hot block cools, the cold one warms, and the flow slows as their temperatures get closer. It stops when both read the same temperature (50 °C for two identical blocks).

Two bodies are in thermal equilibrium when, in contact, no net heat flows between them. Then they have the same temperature, and their measurable properties (volume, pressure, colour, resistance) stop changing.

2The zeroth law

Dip a thermometer C into block A: it reads 37 °C. Dip it into block B: 37 °C again. Now put A and B together: no heat flows. This is the zeroth law:

  • It lets us define temperature: the property that is the same for all bodies in thermal equilibrium.
  • It makes thermometers possible: the thermometer is the third body C.
  • It is called zeroth because it was named after the first and second laws, but it comes before them logically.

3Temperature is not heat

Temperature tells how hot or cold a body is. It is a scalar, and it is equal for bodies in thermal equilibrium. Heat is energy in transit: energy that flows from one body to another because of a temperature difference. A body does not contain heat; it contains internal energy.

4Temperature scales

A scale needs two fixed points: melting ice and boiling water at normal atmospheric pressure.

ScaleIce pointSteam pointSteps between
Celsius (°C)0100100
Fahrenheit (°F)32212180
Kelvin (K)273.15 ≈ 273373.15 ≈ 373100

At any temperature each scale is the same fraction of the way from ice to steam, which gives the conversion rule:

C100=F−32180=K−273100\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}
C=59(F−32)C = \tfrac{5}{9}(F - 32)
F=95C+32F = \tfrac{9}{5}C + 32
K=C+273.15≈C+273K = C + 273.15 \approx C + 273

Celsius and Fahrenheit agree at one temperature: put C=F=xC = F = x, then x=95x+32x = \tfrac{9}{5}x + 32, so x=−40x = -40. So −40 °C = −40 °F (= 233 K).

The kelvin scale starts at absolute zero, 0 K = −273.15 °C ≈ −460 °F, the lowest possible temperature. It can never quite be reached, so a kelvin temperature is never negative. Since 2019 the kelvin is fixed by the Boltzmann constant; the triple point of water is 273.16 K.

5Temperature changes

A kelvin and a Celsius degree are the same size, so a change in temperature is the same number in both. A Fahrenheit degree is only 5/9 as big. For a change, never add 273 or 32.

ΔTK=ΔTC\Delta T_K = \Delta T_C
ΔTF=95 ΔTC\Delta T_F = \tfrac{9}{5}\,\Delta T_C

6Making a thermometer

Any thermometric property XX that changes steadily with temperature will do: the length of a mercury or alcohol column, the pressure of a gas at fixed volume, the resistance of a platinum wire. Mark X0X_0 at the ice point and X100X_{100} at the steam point; then

X−X0X100−X0=T100\frac{X - X_0}{X_{100} - X_0} = \frac{T}{100}T in °C: the same fraction of the way on every scale.

7Mixing to equilibrium

When bodies are mixed and no heat escapes, heat lost = heat gained. The heat capacity C=mcC = mc is the heat a body needs to warm by one degree.

m1c1(T1−T)=m2c2(T−T2)m_1c_1(T_1 - T) = m_2c_2(T - T_2)
T=C1T1+C2T2+…C1+C2+…T = \frac{C_1T_1 + C_2T_2 + \dots}{C_1 + C_2 + \dots}A weighted average: the body with more heat capacity pulls T towards its own temperature.

Summary

Key ideas

  • Bodies in thermal equilibrium have the same temperature and exchange no net heat.
  • By itself, heat flows only from a hotter body to a colder one.
  • Zeroth law: if A and B are each in thermal equilibrium with C, they are in equilibrium with each other.
  • The zeroth law defines temperature and makes thermometers possible (the thermometer is C).
  • Temperature measures hotness; heat is energy in transit; a body holds internal energy.
  • Every scale is the same fraction of the way from the ice point to the steam point.
  • Celsius and Fahrenheit agree at −40.
  • Absolute zero, 0 K = −273.15 °C, cannot be reached; kelvin temperatures are never negative.
  • A change of 1 °C equals a change of 1 K, and of 9/5 °F; never add 273 or 32 to a change.
  • Gas laws need temperatures in kelvin.
  • Mixing: heat lost = heat gained; the final temperature is a heat-capacity-weighted average.

Every equation

Scales
C100=F−32180=K−273100\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}
Celsius
C=59(F−32)C = \tfrac{5}{9}(F - 32)
Fahrenheit
F=95C+32F = \tfrac{9}{5}C + 32
Kelvin
K=C+273.15K = C + 273.15
Absolute zero
0 K=−273.15 ∘C≈−460 ∘F0\ \text{K} = -273.15\ ^\circ\text{C} \approx -460\ ^\circ\text{F}
Same number
−40 ∘C=−40 ∘F-40\ ^\circ\text{C} = -40\ ^\circ\text{F}
Change (K)
ΔTK=ΔTC\Delta T_K = \Delta T_C
Change (°F)
ΔTF=95ΔTC\Delta T_F = \tfrac{9}{5}\Delta T_C
Thermometer
X−X0X100−X0=T100\frac{X - X_0}{X_{100} - X_0} = \frac{T}{100}
Heat lost = gained
m1c1(T1−T)=m2c2(T−T2)m_1c_1(T_1 - T) = m_2c_2(T - T_2)
Mixing
T=∑CiTi∑CiT = \frac{\sum C_iT_i}{\sum C_i}

Previous year questions with solutions

Real JEE and NEET questions on zeroth law of thermodynamics. Try each one before you open the solution.

Q1NEET 2026One correct option

An electric heater supplies heat to a system at a rate of 100 W . If the system performs work at a rate of 75 J/s75\ \mathrm{J/s}, then the rate at which internal energy increases will be:

  1. A75 W
  2. B100 W
  3. C125 W
  4. D25 W
Show answer and solution

Answer: Option D

The first law holds for every second of the process, so it can be written as rates: heat in per second == rise of internal energy per second ++ work done per second.

100=dUdt+75100 = \dfrac{dU}{dt} + 75, so dUdt=25 W\dfrac{dU}{dt} = 25\ \mathrm{W}.

The trap is A, 75 W75\ \mathrm{W}, the rate of doing work, and B, 100 W100\ \mathrm{W}, the whole heating rate — as if no work were done. C, 125 W125\ \mathrm{W}, adds the work to the heat, getting the sign of WW backwards: work done BY the system is energy leaving it.

Q2JEE Main 2026One correct option

Heat is supplied to a diatomic gas at constant pressure. Then the ratio of ΔQ:ΔU:ΔW\Delta Q:\Delta U:\Delta W is ______.

  1. A2 : 3 : 5
  2. B5 : 3 : 2
  3. C2 : 5 : 7
  4. D7 : 5 : 2
Show answer and solution

Answer: Option D

At constant pressure, per mole and per kelvin: ΔQ=CP ΔT\Delta Q = C_{P}\,\Delta T, ΔU=CV ΔT\Delta U = C_{V}\,\Delta T and ΔW=R ΔT\Delta W = R\,\Delta T. For a diatomic gas CV=52RC_{V} = \dfrac{5}{2}R and CP=72RC_{P} = \dfrac{7}{2}R, so

ΔQ:ΔU:ΔW=72:52:1=7:5:2\Delta Q : \Delta U : \Delta W = \dfrac{7}{2} : \dfrac{5}{2} : 1 = 7 : 5 : 2

Check with the first law: 7=5+27 = 5 + 2.

The trap is B, 5:3:25 : 3 : 2, which is the monatomic gas (52:32:1\dfrac{5}{2} : \dfrac{3}{2} : 1). A and C list the same numbers backwards, in the order W:U:QW : U : Q — and neither passes the check that QQ is the sum of the other two.

Q3JEE Advanced 2011Numerical answer

Steel wire of lenght ‘L’ at 40∘40^{\circ}C is suspended from the ceiling and then a mass ‘m’ is hung from its free

end. The wire is cooled down from 40∘40^{\circ}C to 30∘30^{\circ}C to regain its original length ‘L’. The coefficient of linear

thermal expansion of the steel is 10−5 /∘C10^{-5}\ /^{\circ}\mathrm{C}, Young’s modulus of steel is 10¹¹ N/m² and radius of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of ‘m’ in kg is nearly

Show answer and solution

Answer: 3

The load stretches the wire by mgLAY\dfrac{mgL}{AY}; cooling by 10∘C10^{\circ}\mathrm{C} shrinks it by Lα ΔTL\alpha\,\Delta T. The wire is back to LL when the two are equal: mgπr2Y=α ΔT\dfrac{mg}{\pi r^{2}Y} = \alpha\,\Delta T, so m=πr2Yα ΔTg=3.14×10−6×1011×10−5×109.8≈3.2 kgm = \dfrac{\pi r^{2}Y\alpha\,\Delta T}{g} = \dfrac{3.14 \times 10^{-6} \times 10^{11} \times 10^{-5} \times 10}{9.8} \approx 3.2\ \mathrm{kg}, which is nearly 3 kg3\ \mathrm{kg}. The length LL cancels.

The trap is using the diameter, 2 mm2\ \mathrm{mm}, in place of the radius, which gives four times too much, about 13 kg13\ \mathrm{kg}. Using the whole 40∘C40^{\circ}\mathrm{C} in place of the 1010-degree fall also gives four times too much.

Practice questions, easy to hard

Three questions from the zeroth law of thermodynamics practice ladder: one easy, one medium, one hard.

Q4One correct option

A thermometer with a faulty scale is just another scale X. One that reads 55 in melting ice and 9595 in boiling water has Xi=5X_{i} = 5 and Xs=95X_{s} = 95: 9090 of its divisions cover what 100100 Celsius degrees do.

In a bath, this faulty thermometer reads 3232. What is the bath's true temperature?

  1. A27∘C27^{\circ}\mathrm{C}
  2. B30∘C30^{\circ}\mathrm{C}
  3. C35.6∘C35.6^{\circ}\mathrm{C}
  4. D32∘C32^{\circ}\mathrm{C}
Show answer and solution

Answer: Option B

C100=32−595−5=2790=0.3\dfrac{C}{100} = \dfrac{32 - 5}{95 - 5} = \dfrac{27}{90} = 0.3, so C=30∘CC = 30^{\circ}\mathrm{C}.

The trap is A, 27∘C27^{\circ}\mathrm{C}: it takes off the 55 but forgets that each faulty division is bigger than a Celsius degree. C, 35.6∘C35.6^{\circ}\mathrm{C}, scales by 10090\dfrac{100}{90} but forgets to start from the ice point, and D simply trusts the faulty reading. Check: 30∘C30^{\circ}\mathrm{C} is 0.30.3 of the way up, and 5+0.3×90=325 + 0.3 \times 90 = 32.

Q5One or more correct options

A liquid has no shape of its own, so it has no length or area to speak of: only volume expansion counts, with the liquid's own γ\gamma. It also has to be kept in a vessel, and the vessel expands too — its inside grows as if it were filled with the vessel's material, with the vessel's γ\gamma. So what we see is the apparent expansion of the liquid, relative to the vessel:

γapp=γliquid−γvessel\gamma_{\mathrm{app}} = \gamma_{\mathrm{liquid}} - \gamma_{\mathrm{vessel}}

A vessel filled to the brim with a volume V0V_{0} of liquid and warmed by ΔT\Delta T overflows by V0 γapp ΔTV_{0}\,\gamma_{\mathrm{app}}\,\Delta T, not by the liquid's whole expansion.

A glass flask is filled to the brim with a liquid whose γ\gamma is larger than that of the glass, and both are warmed together. Which statements are correct?

  1. ASome of the liquid overflows
  2. BThe volume that overflows is the liquid's whole increase in volume
  3. CHad the liquid's γ\gamma equalled that of the glass, none would overflow, and the flask would stay just full
  4. DThe inside of the flask grows by V0 γglass ΔTV_{0}\,\gamma_{\mathrm{glass}}\,\Delta T, as if it were solid glass
Show answer and solution

Answer: Options A, C, D

A: the liquid grows faster than the space holding it, so the excess spills. C: with equal coefficients liquid and space grow alike, γapp=0\gamma_{\mathrm{app}} = 0. D is the cavity rule: the inside of the flask grows as the glass that would fill it.

B is the trap. The flask has made room for part of the liquid's growth, V0 γglass ΔTV_{0}\,\gamma_{\mathrm{glass}}\,\Delta T, so only the rest, V0(γliquid−γglass) ΔTV_{0}(\gamma_{\mathrm{liquid}} - \gamma_{\mathrm{glass}})\,\Delta T, spills out. The overflow measures the apparent expansion, not the real one.

Q6One correct option

A bimetallic strip bends into a curve of radius R≈d(α1−α2) ΔTR \approx \dfrac{d}{(\alpha_{1} - \alpha_{2})\,\Delta T}, where dd is the thickness of each strip and ΔT\Delta T the change from the temperature at which it was straight.

A strip is made of brass (α=1.9×10−5 K−1\alpha = 1.9 \times 10^{-5}\ \mathrm{K^{-1}}) and iron (α=1.2×10−5 K−1\alpha = 1.2 \times 10^{-5}\ \mathrm{K^{-1}}), each 0.50 mm0.50\ \mathrm{mm} thick, and is straight at 20∘C20^{\circ}\mathrm{C}. What is the radius of the curve it bends into at 120∘C120^{\circ}\mathrm{C}?

  1. A7.1 m7.1\ \mathrm{m}
  2. B1.4 m1.4\ \mathrm{m}
  3. C0.60 m0.60\ \mathrm{m}
  4. D0.71 m0.71\ \mathrm{m}
Show answer and solution

Answer: Option D

R≈0.50×10−3(1.9−1.2)×10−5×100=5.0×10−47.0×10−4≈0.71 mR \approx \dfrac{0.50 \times 10^{-3}}{(1.9 - 1.2) \times 10^{-5} \times 100} = \dfrac{5.0 \times 10^{-4}}{7.0 \times 10^{-4}} \approx 0.71\ \mathrm{m}.

The trap is C, 0.60 m0.60\ \mathrm{m}, which uses 120120 as the change: the strip was straight at 20∘C20^{\circ}\mathrm{C}, so ΔT=100 K\Delta T = 100\ \mathrm{K}. B, 1.4 m1.4\ \mathrm{m}, uses the whole strip's thickness, 1.0 mm1.0\ \mathrm{mm}, for dd, which is the thickness of each strip. A, 7.1 m7.1\ \mathrm{m}, slips a factor of ten in turning millimetres into metres.