JEE MainPhysics

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+4 marks−1 if incorrectSingle correctPrevious-year question

How to apply Newton's Law of Cooling to successive time intervals

Q: Why do we take the average temperature (T_1+T_2)/2 instead of integrating?

The average approximation dT/dt ≈ (T_1-T_2)/t = K(T_avg - T_s) is standard in JEE Main for moderate temperature differences and gives nearly identical results to the exponential decay formula with simpler arithmetic.

A body cools in 7minutes from 60°C to 40°C. The temperature of the surrounding is 10°C. The temperature of the body after the next 7minutes will be:

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A32°CB30°CC28°CD34°C

Step-by-step solutionView

Correct answer

Using Newton's law of cooling, the temperature difference from the surroundings decays by a factor of 0.6 every 7 minutes, resulting in a temperature of 28°C after the second interval.

Option analysis

Why each option works or fails

A · 32°C

Believing that average temperature should be calculated without dividing the sum of initial and final temperatures by two. When using the approximate form of Newton's law of cooling, ensure the body's mean temperature is computed as (T₁ + T₂)/2 before subtracting the ambient temperature.

B · 30°C

Assuming that the rate of temperature decrease is constant over equal time intervals, leading to a linear drop of 20°C in each 7-minute period, or misplacing the ambient temperature reference. The rate of cooling is proportional to the temperature excess over the environment, which decreases as the body cools; the temperature drop in the second interval must be strictly less than in the first.

C · 28°C

The exponential decay of temperature excess (or its standard linear approximation) is correctly applied across both intervals. This is the correct answer.

D · 34°C

Making an arithmetic error when solving the algebraic equation for the final temperature. Carefully solve the resulting equation after setting up the cooling constant equivalence between the two intervals.

Reviewed route

Solution

StepWorking

01Given

Initial temperature $T_{0} = 60^{\circ} C$ , after $t_{1} = 7 min$ temperature $T_{1} = 40^{\circ} C$ , surrounding temperature $T_{s} = 10^{\circ} C$ , next interval $t_{2} = 7 min$ .

02Find

Find the temperature $T_{2}$ of the body after the next $7 minutes$ .

03Visualise

A body is cooling in a constant surrounding temperature of $10^{\circ} C$ . In the first $7 min$ , it drops by $20^{\circ} C$ (from $60^{\circ} C$ to $40^{\circ} C$ ). As the temperature difference with the surroundings decreases, the rate of cooling slows down, so in the next $7 min$ it will cool by less than $20^{\circ} C$ .

04Strategise

Apply the average form of Newton's law of cooling: $\frac{T_{i} - T_{f}}{t} = K (\frac{T_{i} + T_{f}}{2} - T_{s})$ . Form two equations for the two consecutive 7-minute intervals and solve for $T_{2}$ .

05Execute

For interval 1: $\frac{60 - 40}{7} = K (50 - 10) ⟹ \frac{20}{7} = 40 K ⟹ K = \frac{1}{14} {min}^{- 1}$ . For interval 2: $\frac{40 - T_{2}}{7} = \frac{1}{14} (\frac{40 + T_{2}}{2} - 10) ⟹ 40 - T_{2} = \frac{1}{2} (\frac{20 + T_{2}}{2}) = \frac{20 + T_{2}}{4}$ . Multiplying both sides by 4: $160 - 4 T_{2} = 20 + T_{2} ⟹ 5 T_{2} = 140 ⟹ T_{2} = 28^{\circ} C$ .

✓Verify

The temperature drop in the second interval is $40 - 28 = 12^{\circ} C$ , which is less than the first drop of $20^{\circ} C$ . This is physically consistent with Newton's law of cooling where cooling rate decreases with decreasing $Δ T$ .

Hints that build this answer step by step

Which relationship correctly describes how the excess temperature $θ = T - T_{0}$ changes with time under Newton's Law of Cooling?
$θ (t) = θ (0) e^{- k t}$ , meaning the ratio of temperature excess is constant over equal time intervals
What is the decay factor for the excess temperature $θ = T - 10^{\circ} C$ over the first $7 minutes$ ?
$\frac{θ_{1}}{θ_{0}} = \frac{40 - 10}{60 - 10} = \frac{30}{50} = 0.6$
Using this decay factor of $0.6$ , what will be the excess temperature and final temperature after the next $7 minutes$ ?
$θ_{2} = 0.6 \times 30 = 18^{\circ} C$ , so $T_{2} = 18 + 10 = 28^{\circ} C$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Why do we take the average temperature

(T_{1} + T_{2}) / 2

instead of integrating?

The average approximation $\frac{d T}{d t} \approx \frac{T_{1} - T_{2}}{t} = K (T_{a v g} - T_{s})$ is standard in JEE Main for moderate temperature differences and gives nearly identical results to the exponential decay formula with simpler arithmetic.

Answer

Using Newton's law of cooling, the temperature difference from the surroundings decays by a factor of 0.6 every 7 minutes, resulting in a temperature of 28°C after the second interval.

Why each option works or fails

A: 32°C - Believing that average temperature should be calculated without dividing the sum of initial and final temperatures by two. When using the approximate form of Newton's law of cooling, ensure the body's mean temperature is computed as (T₁ + T₂)/2 before subtracting the ambient temperature.
B: 30°C - Assuming that the rate of temperature decrease is constant over equal time intervals, leading to a linear drop of 20°C in each 7-minute period, or misplacing the ambient temperature reference. The rate of cooling is proportional to the temperature excess over the environment, which decreases as the body cools; the temperature drop in the second interval must be strictly less than in the first.
C · correct: 28°C - The exponential decay of temperature excess (or its standard linear approximation) is correctly applied across both intervals. This is the correct answer.
D: 34°C - Making an arithmetic error when solving the algebraic equation for the final temperature. Carefully solve the resulting equation after setting up the cooling constant equivalence between the two intervals.

Step-by-step solution

Given: Initial temperature $T_{0} = 60^{\circ} C$ , after $t_{1} = 7 min$ temperature $T_{1} = 40^{\circ} C$ , surrounding temperature $T_{s} = 10^{\circ} C$ , next interval $t_{2} = 7 min$ .
Find: Find the temperature $T_{2}$ of the body after the next $7 minutes$ .
Visualise: A body is cooling in a constant surrounding temperature of $10^{\circ} C$ . In the first $7 min$ , it drops by $20^{\circ} C$ (from $60^{\circ} C$ to $40^{\circ} C$ ). As the temperature difference with the surroundings decreases, the rate of cooling slows down, so in the next $T_{0} = 60^{\circ} C$ it will cool by less than $20^{\circ} C$ .
Strategise: Apply the average form of Newton's law of cooling: $\frac{T_{i} - T_{f}}{t} = K (\frac{T_{i} + T_{f}}{2} - T_{s})$ . Form two equations for the two consecutive 7-minute intervals and solve for $T_{2}$ .
Execute: For interval 1: $\frac{60 - 40}{7} = K (50 - 10) ⟹ \frac{20}{7} = 40 K ⟹ K = \frac{1}{14} {min}^{- 1}$ . For interval 2: $\frac{40 - T_{2}}{7} = \frac{1}{14} (\frac{40 + T_{2}}{2} - 10) ⟹ 40 - T_{2} = \frac{1}{2} (\frac{20 + T_{2}}{2}) = \frac{20 + T_{2}}{4}$ . Multiplying both sides by 4: $160 - 4 T_{2} = 20 + T_{2} ⟹ 5 T_{2} = 140 ⟹ T_{2} = 28^{\circ} C$ .
Verify: The temperature drop in the second interval is $40 - 28 = 12^{\circ} C$ , which is less than the first drop of $20^{\circ} C$ . This is physically consistent with Newton's law of cooling where cooling rate decreases with decreasing $Δ T$ .

How to apply Newton's Law of Cooling to successive time intervals

What feels right?

Using Newton's law of cooling, the temperature difference from the surroundings decays by a factor of 0.6 every 7 minutes, resulting in a temperature of 28°C after the second interval.

Why each option works or fails

Solution

01Given

02Find

03Visualise

04Strategise

05Execute

✓Verify

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Using Newton's law of cooling, the temperature difference from the surroundings decays by a factor of 0.6 every 7 minutes, resulting in a temperature of 28°C after the second interval.

01Given

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03Visualise

04Strategise

05Execute

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