Properties of Solids and Liquids: Physics | JEE Main
A body cools in 7 minutes from 60∘C to 40∘C. The temperature of the surrounding is 10∘C. The temperature of the body after the next 7 minutes will be:
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Hint 1 of 3
Which relationship correctly describes how the excess temperature θ=T−T0 changes with time under Newton's Law of Cooling?
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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
None. 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 T0=60∘C, after t1=7 min temperature T1=40∘C, surrounding temperature Ts=10∘C, next interval t2=7 min.
02find
Find the temperature T2 of the body after the next 7 minutes.
03visualise
A body is cooling in a constant surrounding temperature of 10∘C. In the first 7 min, it drops by 20∘C (from 60∘C to 40∘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∘C.
04strategise
Apply the average form of Newton's law of cooling: tTi−Tf=K(2Ti+Tf−Ts). Form two equations for the two consecutive 7-minute intervals and solve for T2.
05execute
For interval 1: 760−40=K(50−10)⟹720=40K⟹K=141 min−1.
For interval 2: 740−T2=141(240+T2−10)⟹40−T2=21(220+T2)=420+T2.
Multiplying both sides by 4: 160−4T2=20+T2⟹5T2=140⟹T2=28∘C.
✓verify
The temperature drop in the second interval is 40−28=12∘C, which is less than the first drop of 20∘C. This is physically consistent with Newton's law of cooling where cooling rate decreases with decreasing ΔT.
Hints that build this answer step by step
Which relationship correctly describes how the excess temperature θ=T−T0 changes with time under Newton's Law of Cooling?
θ(t)=θ(0)e−kt, meaning the ratio of temperature excess is constant over equal time intervals
What is the decay factor for the excess temperature θ=T−10∘C over the first 7 minutes?
θ0θ1=60−1040−10=5030=0.6
Using this decay factor of 0.6, what will be the excess temperature and final temperature after the next 7 minutes?
Why do we take the average temperature (T1+T2)/2 instead of integrating?
The average approximation dtdT≈tT1−T2=K(Tavg−Ts) is standard in JEE Main for moderate temperature differences and gives nearly identical results to the exponential decay formula with simpler arithmetic.