Concept of Temperature

8 MCQs9-step worked example

Source: NCERT ThermodynamicsPYQ coverage: NEET 2022, 2023, 2024, 2025Official key: NTA-verifiedLast reviewed: May 2026

Lesson

Topic: Concept of Temperature

Temperature is one of the most familiar physical quantities, yet NEET aspirants often stumble on its precise thermodynamic definition. The core idea: temperature determines the direction of heat flow between two systems in thermal contact. Heat flows from higher temperature to lower temperature — and when it stops, the systems are in thermal equilibrium.

The Zeroth Law of Thermodynamics (NCERT Class 11 Physics, Chapter 11, page 2) formalises this: if system A is in thermal equilibrium with system C, and system B is also in thermal equilibrium with system C, then A and B are in thermal equilibrium with each other. This transitive property is what makes thermometers meaningful — system C acts as the thermometer.

Temperature is a scalar state function. It depends only on the current state of the system, not how it got there. For an ideal gas, temperature is directly proportional to the average translational kinetic energy of molecules: ⟨KE⟩ = (3/2)kT, where k is Boltzmann's constant.

Temperature scales and conversions appear regularly in NEET:

Celsius to Kelvin: T(K) = T(°C) + 273.15
Fahrenheit to Celsius: T(°C) = (5/9)[T(°F) − 32]
The Kelvin scale is the SI thermodynamic scale; 0 K is absolute zero.

A common confusion: students treat temperature as a measure of "total heat" in a body. Temperature measures average molecular kinetic energy per degree of freedom — two bodies at the same temperature can hold vastly different amounts of thermal energy depending on mass and specific heat.

Watch out for conversion errors in numerical problems — forgetting the 273.15 offset or misapplying the (5/9) vs (9/5) factor in Fahrenheit conversions costs easy marks.

Practice MCQs

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The Zeroth Law of Thermodynamics is the basis for the concept of:

MCQ 2Easy RecallPractice

What is the SI unit of temperature?

MCQ 3Easy RecallPractice

If system A is in thermal equilibrium with system C, and system B is in thermal equilibrium with system C, then according to the Zeroth Law:

MCQ 4Direct ApplicationPractice

Convert 37°C (normal human body temperature) to Kelvin.

MCQ 5Direct ApplicationPractice

A temperature of −40°F is equivalent to what value on the Celsius scale?

MCQ 6Direct ApplicationPractice

Two bodies A and B are at temperatures 300 K and 400 K respectively. When placed in thermal contact, heat will flow:

MCQ 7Concept TrapPractice

A large block of iron and a small cup of water are both at 80°C. Which statement is correct?

MCQ 8CalculationPractice

A thermometer reads 50°F. Convert this to the Kelvin scale.

Worked Example

1
Given
A thermometer reads 212°F.
2
Required
Express this temperature in (a) Celsius and (b) Kelvin.
3
Concept
Temperature scale conversion uses the Fahrenheit-to-Celsius formula and the Celsius-to-Kelvin offset. These are direct applications of the definitions of the three common temperature scales.
4
Formula
- T(°C) = (5/9)[T(°F) − 32] - T(K) = T(°C) + 273.15
5
Substitution
- T(°C) = (5/9)(212 − 32) = (5/9)(180) - T(K) = T(°C) + 273.15
6
Calculation
- T(°C) = (5/9) × 180 = 100°C - T(K) = 100 + 273.15 = 373.15 K Note on exact values: 32 and 273.15 are defined conversion constants (exact by definition). The input 212°F is the exact boiling point of water at 1 atm. These do not introduce significant-figure limitations.
7
Final answer
212°F = 100°C = 373.15 K
8
Common trap
The most common error is using (9/5) instead of (5/9) when converting from Fahrenheit to Celsius — that would give T(°C) = (9/5)(180) = 324°C, a wildly incorrect answer. Remember: Fahrenheit-to-Celsius uses the fraction 5/9; Celsius-to-Fahrenheit uses 9/5. A second trap: forgetting to subtract 32 first. If you compute (5/9)(212) = 117.8°C, you get a wrong intermediate value.
9
Similar NEET-style question
"The temperature of a furnace is 1832°F. What is the temperature in Kelvin?" (Answer: T(°C) = (5/9)(1832 − 32) = (5/9)(1800) = 1000°C → T(K) = 1273.15 K) ---

Before solving, remember these

Law

Zeroth law of thermodynamics

If two systems are each in thermal equilibrium with a third, they are in thermal equilibrium with each other. Defines temperature operationally.

-- NCERT, p. 2

Formulas

4 formulas — click to collapse

Adiabatic relations for ideal gas

P V^{γ} = const; T V^{γ - 1} = const

Relations holding during reversible adiabatic process. gamma = Cp/Cv.

Symbol	Quantity	SI Unit
P	pressure	Pa
V	volume	m^3
T	temperature	K
gamma	adiabatic index	-

Valid when

Q = 0 (no heat exchange)
Quasi-static (reversible)
Ideal gas

First law of thermodynamics

Δ U = Q - W

Change in internal energy = heat ADDED minus work DONE BY the system. Energy conservation including thermal energy.

Symbol	Quantity	SI Unit
Delta_U	change in internal energy	J
Q	heat added to system	J
W	work done BY system	J

Valid when

Closed system (no mass exchange)
Sign convention: Q>0 heat in, W>0 system does work

Work done in isothermal process (ideal gas)

W = n R T ln \frac{V _{f}}{V _{i}}

Work done by ideal gas during isothermal expansion. Q = W (since Delta_U = 0). Reverse for compression.

Symbol	Quantity	SI Unit
n	moles	mol
R	gas constant 8.314	J/mol/K
T	temperature	K
V_i, V_f	initial/final volume	m^3

Valid when

Ideal gas
Quasi-static (reversible) isothermal

Mayer's relation (Cp - Cv = R)

C_{p} - C_{v} = R

For ideal gas: difference of molar specific heats equals gas constant R. Useful for converting between Cp and Cv.

Symbol	Quantity	SI Unit
Cp	molar specific heat at const P	J/mol/K
Cv	molar specific heat at const V	J/mol/K
R	8.314	J/mol/K

Valid when

Ideal gas
Per mole basis

Exam Traps & Common Mistakes

These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.

2 items — click to collapse

Trap

Adiabatic vs isothermal P-V curve

Category: Graph Interpretation

Student misidentifies which P-V curve is adiabatic (steeper) vs isothermal.

When it triggers

P-V graph showing one or more processes; question asks for process type.

How to avoid

Adiabatic curve is STEEPER than isothermal at the same point (slope ratio = γ). Adiabat: PV^γ; isotherm: PV = const.

Mistake

Identifies adiabatic curve as gentler than isothermal on P-V diagram (or vice versa).

Root cause: graph misread

Correction

Adiabatic curve is STEEPER than isothermal at the same P-V point. Slope ratio at same point: adiabatic / isothermal = γ. Memorise: 'adiabatic angles down sharper'.