Zeroth Law of Thermodynamics

Master thermal equilibrium and temperature scales for JEE Main & Advanced with real-world applications

9 min read

Real-Life Hook: Why Do Metal Benches Feel Colder?

Ever sat on a metal bench in winter and jumped up immediately? The metal isn’t actually colder than a wooden bench beside it—they’re both at the same temperature! This everyday mystery reveals the foundation of thermodynamics: thermal equilibrium and how we actually measure temperature.

When you touch metal, it conducts heat away from your skin faster than wood does, making it feel colder. But given enough time, your thermometer would read the same temperature for both. This is the Zeroth Law in action!

The Zeroth Law of Thermodynamics

Statement

If two systems A and B are separately in thermal equilibrium with a third system C, then A and B are in thermal equilibrium with each other.

This seemingly obvious statement is the foundation for temperature measurement. System C is typically a thermometer!

What is Thermal Equilibrium?

Two systems are in thermal equilibrium when:

  • No net heat flows between them when brought in contact
  • They have the same temperature
  • Their macroscopic properties stop changing

Temperature: The Fundamental Concept

Temperature is NOT:

  • ❌ Heat energy
  • ❌ Average kinetic energy (that’s for ideal gases specifically)

Temperature IS:

  • ✅ A measure of the degree of hotness or coldness
  • ✅ The property that’s equal for systems in thermal equilibrium
  • ✅ A scalar quantity with direction of heat flow

Heat Flow Rule

Heat always flows from higher temperature to lower temperature until thermal equilibrium is reached.

Temperature Scales

Celsius Scale (°C)

Reference Points:

  • Ice point: 0°C (freezing point of water at 1 atm)
  • Steam point: 100°C (boiling point of water at 1 atm)
  • 100 equal divisions between these points

Fahrenheit Scale (°F)

Reference Points:

  • Ice point: 32°F
  • Steam point: 212°F
  • 180 equal divisions between these points

Kelvin Scale (K) - The Absolute Scale

Reference Points:

  • Absolute zero: 0 K (lowest possible temperature)
  • Triple point of water: 273.16 K (exact definition)
  • Ice point: 273.15 K ≈ 273 K (for JEE calculations)

Why Kelvin? It’s the only scale where temperature is directly proportional to average kinetic energy of molecules. No negative values!

Temperature Conversion Formulas

Key Conversions:

$$\frac{C}{100} = \frac{F - 32}{180} = \frac{K - 273}{100}$$

Simplified versions:

$$C = \frac{5}{9}(F - 32)$$ $$F = \frac{9}{5}C + 32$$ $$K = C + 273$$

For temperature differences:

$$\Delta C = \Delta K$$ $$\Delta F = \frac{9}{5}\Delta C = \frac{9}{5}\Delta K$$

Memory Tricks

“FBI” Trick for Conversion

  • Fahrenheit uses Fractions with 9 on top: 9/5
  • Both Celsius and Kelvin have Base 100 divisions
  • Ice point differences: F has +32, K has +273

Temperature Difference Trick

“Change in Celsius = Change in Kelvin, but Change in Fahrenheit is 9/5 bigger”

$$\Delta T_{°C} = \Delta T_K \neq \Delta T_{°F}$$

Absolute Zero Memory

  • Celsius: -273°C
  • Fahrenheit: -460°F
  • Kelvin: 0 K (by definition)

Mnemonic: “Celsius to Kelvin: Just add TWO SEVEN THREE”

Common Mistakes to Avoid

Mistake 1: Using °C instead of K in Gas Laws

❌ Wrong: $PV = nRT$ with $T = 27°C$ ✅ Correct: $PV = nRT$ with $T = 300 K$

Rule: Always convert to Kelvin for thermodynamic calculations!

Mistake 2: Confusing Temperature and Temperature Difference

  • Temperature: $T_K = T_C + 273$
  • Temperature difference: $\Delta T_K = \Delta T_C$ (no +273!)

Mistake 3: Saying “Metal is Colder”

Objects in the same room are at the same temperature. Metal just has higher thermal conductivity (connects to heat transfer, not thermodynamics).

Mistake 4: Negative Kelvin

Kelvin can NEVER be negative. If you get negative Kelvin, you made a calculation error!

JEE-Style Concept Applications

Type 1: Direct Conversion Problems

Example: A temperature of 40°F is equivalent to:

  1. Find Celsius: $C = \frac{5}{9}(40 - 32) = \frac{5}{9}(8) = 4.44°C$
  2. Find Kelvin: $K = 4.44 + 273 = 277.44 K$

Type 2: Temperature Difference Problems

Example: A metal rod expands when heated from 20°C to 100°C. Express this change in Kelvin and Fahrenheit.

  • $\Delta T_C = 100 - 20 = 80°C$
  • $\Delta T_K = 80 K$ (same magnitude!)
  • $\Delta T_F = \frac{9}{5} \times 80 = 144°F$

Type 3: Thermal Equilibrium

Example: Three bodies A, B, C are at temperatures 300 K, 400 K, and 300 K respectively. Which are in thermal equilibrium?

Only A and C (same temperature = thermal equilibrium).

Practice Problems

Level 1: JEE Main Basics

Q1. The temperature of a system rises by 90°C. Express this rise in Kelvin and Fahrenheit.

Solution
  • Rise in Kelvin: $\Delta T_K = \Delta T_C = 90 K$
  • Rise in Fahrenheit: $\Delta T_F = \frac{9}{5} \times 90 = 162°F$

Key Point: For temperature changes, C and K are equal!

Q2. At what temperature do the Celsius and Fahrenheit scales give the same numerical value?

Solution

Let the temperature be $x$ on both scales:

$$x = \frac{9}{5}x + 32$$ $$5x = 9x + 160$$ $$-4x = 160$$ $$x = -40$$

So -40°C = -40°F

Memory Tip: This is the only “same number” point for C and F!

Q3. Which of the following statements about the Zeroth Law is correct?

  1. It defines temperature
  2. It establishes thermal equilibrium as transitive
  3. It allows temperature measurement
  4. All of the above
Solution

Answer: 4. All of the above

The Zeroth Law:

  • Defines temperature as the property equal in thermal equilibrium
  • Shows transitivity: A = C and B = C → A = B
  • Makes thermometers possible (thermometer is the third system C)

Level 2: JEE Main/Advanced

Q4. Two ideal gases are at temperatures $T_1$ and $T_2$. When mixed in a calorimeter, the final temperature is 400 K. If $T_1 = 300 K$ and the gases have equal masses and specific heats, find $T_2$.

Solution

For equal masses and specific heats, heat lost = heat gained:

$$m c (T_2 - 400) = m c (400 - 300)$$ $$T_2 - 400 = 100$$ $$T_2 = 500 K$$

Alternative: For equal heat capacities, final temperature is arithmetic mean:

$$T_f = \frac{T_1 + T_2}{2}$$ $$400 = \frac{300 + T_2}{2}$$ $$T_2 = 500 K$$

Q5. A faulty thermometer reads 5°C in melting ice and 105°C in steam. What is the correct temperature when this thermometer reads 50°C?

Solution

The faulty thermometer has:

  • Lower fixed point: 5°C (should be 0°C)
  • Upper fixed point: 105°C (should be 100°C)
  • Range: 100 divisions

When reading shows 50°C:

  • Reading above lower point: 50 - 5 = 45
  • Fraction of scale: 45/100

Actual temperature:

$$T = 0 + \frac{45}{100} \times 100 = 45°C$$

Formula Method:

$$\frac{T - 0}{100 - 0} = \frac{50 - 5}{105 - 5}$$ $$\frac{T}{100} = \frac{45}{100}$$ $$T = 45°C$$

Level 3: JEE Advanced

Q6. A platinum resistance thermometer has resistance 3.00 Ω at 0°C and 3.75 Ω at 100°C. The thermometer is immersed in a liquid and shows resistance 3.15 Ω. Find the temperature of the liquid. Also express in Kelvin and Fahrenheit.

Solution

Assuming linear variation:

$$\frac{R - R_0}{R_{100} - R_0} = \frac{T - 0}{100 - 0}$$ $$\frac{3.15 - 3.00}{3.75 - 3.00} = \frac{T}{100}$$ $$\frac{0.15}{0.75} = \frac{T}{100}$$ $$T = \frac{0.15 \times 100}{0.75} = 20°C$$

In Kelvin: $T = 20 + 273 = 293 K$

In Fahrenheit: $F = \frac{9}{5}(20) + 32 = 36 + 32 = 68°F$

Q7. Three bodies A, B, and C have heat capacities $C_1$, $2C_1$, and $3C_1$ respectively and are at temperatures 100°C, 50°C, and 20°C. If A and B are brought in contact until equilibrium, and then C is added, find the final temperature.

Solution

Step 1: A and B reach equilibrium

$$C_1(100 - T_{AB}) = 2C_1(T_{AB} - 50)$$ $$100 - T_{AB} = 2T_{AB} - 100$$ $$200 = 3T_{AB}$$ $$T_{AB} = \frac{200}{3}°C$$

Step 2: System AB (now at 200/3°C) with C Combined heat capacity of AB: $C_1 + 2C_1 = 3C_1$

$$3C_1\left(\frac{200}{3} - T_f\right) = 3C_1(T_f - 20)$$ $$\frac{200}{3} - T_f = T_f - 20$$ $$\frac{200}{3} + 20 = 2T_f$$ $$\frac{200 + 60}{3} = 2T_f$$ $$T_f = \frac{260}{6} = \frac{130}{3} ≈ 43.33°C$$

Alternative (Direct Method):

$$T_f = \frac{C_1(100) + 2C_1(50) + 3C_1(20)}{C_1 + 2C_1 + 3C_1} = \frac{100 + 100 + 60}{6} = \frac{260}{6} = 43.33°C$$

Q8. The coefficient of linear expansion of a metal is $\alpha$ per °C. A metal rod of length $L_0$ at 0°C is heated. Show that the relation between length and temperature is the same whether temperature is measured in Celsius or Kelvin.

Solution

In Celsius:

$$L = L_0(1 + \alpha T_C)$$

In Kelvin: Since $T_K = T_C + 273$, we need to measure from a reference.

If we measure expansion from 0°C (= 273 K):

$$L = L_0(1 + \alpha (T_K - 273))$$ $$L = L_0(1 + \alpha T_K - 273\alpha)$$

But this looks different! The resolution:

The coefficient $\alpha$ per °C is the same as per K because:

$$\Delta T_C = \Delta T_K$$

So when we write:

$$L = L_0[1 + \alpha(T - T_0)]$$

Whether we use $T$ in °C or K, the difference $(T - T_0)$ is the same:

$$L = L_0[1 + \alpha(T_C - 0)] = L_0[1 + \alpha(T_K - 273)]$$

Both give the same length!

Key Insight: Expansion depends on change in temperature, and $\Delta T_C = \Delta T_K$.

Connection to Other Topics

→ Kinetic Theory of Gases

Temperature in Kelvin is directly proportional to average kinetic energy:

$$\frac{1}{2}m\langle v^2 \rangle = \frac{3}{2}k_B T$$

This is why we MUST use Kelvin in gas laws!

Learn more in Kinetic Theory →

→ Calorimetry

Thermal equilibrium is the basis of calorimetry:

  • Mixed substances reach common final temperature
  • Heat lost = Heat gained (if no heat loss to surroundings)

→ Heat Transfer

  • Conduction: Why metal feels colder (thermal conductivity)
  • Radiation: Stefan-Boltzmann law uses absolute temperature (K)

→ Chemistry: Thermodynamics

  • Standard temperature: 273 K or 298 K
  • Gibbs free energy: Uses absolute temperature
  • Equilibrium constants: Temperature in Kelvin

See Chemistry Thermodynamics →

Quick Revision Points

  1. Zeroth Law: If A = C and B = C, then A = B (makes thermometers possible)
  2. Temperature ≠ Heat: Temperature measures “hotness,” heat is energy transfer
  3. Kelvin is King: Always use K in thermodynamic equations
  4. Conversion: $K = C + 273$, $F = \frac{9}{5}C + 32$
  5. Changes: $\Delta C = \Delta K$, but $\Delta F = \frac{9}{5}\Delta K$
  6. Thermal Equilibrium: No net heat flow, same temperature
  7. Heat Flow: High T → Low T (never reversed spontaneously)
  8. Absolute Zero: 0 K = -273°C = -460°F (unattainable in practice)

JEE Previous Year Insights

  • Temperature conversion: Asked in 40% of thermodynamics problems
  • Faulty thermometer: Classic JEE Main question (appears every 2-3 years)
  • Thermal equilibrium with mixing: Common in JEE Advanced
  • Always check: Is temperature in Kelvin? Most common error!

Next Topic: First Law of Thermodynamics →

Last updated: February 2025