Newton's law of cooling — body cools from 70°C to 60°C in 5 min — time to reach 50°C

Question

A body cools from 70°C to 60°C in 5 minutes. The temperature of the surroundings is 30°C. How much time will the body take to cool from 60°C to 50°C?

Solution — Step by Step

Newton’s Law of Cooling states that the rate of heat loss of a body is proportional to the difference between the body’s temperature and the surrounding temperature:

\frac{dT}{dt} = -k(T - T_s)

where $T$ = temperature of body, $T_s$ = temperature of surroundings, $k$ = cooling constant (positive, depends on object’s properties).

Integrating this: $T(t) = T_s + (T_0 - T_s)e^{-kt}$ — but for CBSE/JEE purposes, an approximate form is more commonly used for short time intervals.

For small time intervals, Newton’s Law of Cooling can be approximated as:

\frac{\Delta T}{\Delta t} \approx -k\left(\frac{T_1 + T_2}{2} - T_s\right)

where $T_1$ and $T_2$ are the temperatures at the start and end of the interval, $\Delta T = T_2 - T_1$ , and $\Delta t$ is the time interval.

This approximation uses the average temperature during the interval.

$T_1 = 70°C$ , $T_2 = 60°C$ , $\Delta T = -10°C$ , $\Delta t = 5$ min, $T_s = 30°C$ .

Average temperature: $T_{avg1} = \frac{70 + 60}{2} = 65°C$

Applying the approximate law:

\frac{-10}{5} = -k(65 - 30)

-2 = -k \times 35

k = \frac{2}{35} \text{ min}^{-1}

$T_1 = 60°C$ , $T_2 = 50°C$ , $\Delta T = -10°C$ , time = $t_2$ (unknown), $T_s = 30°C$ .

Average temperature: $T_{avg2} = \frac{60 + 50}{2} = 55°C$

Applying the same law with $k = 2/35$ :

\frac{-10}{t_2} = -\frac{2}{35}(55 - 30)

\frac{10}{t_2} = \frac{2}{35} \times 25 = \frac{50}{35} = \frac{10}{7}

t_2 = \frac{10}{\frac{10}{7}} = 7 \text{ min}

The body takes 7 minutes to cool from 60°C to 50°C.

Physical interpretation: Even though the temperature drop is the same (10°C in both cases), the second interval takes longer. This makes sense — as the body approaches ambient temperature, the temperature difference driving heat loss is smaller, so cooling slows down.

Why This Works

Newton’s Law of Cooling is essentially saying that heat flows faster when the temperature gradient is larger. As the body cools, the driving force (temperature difference between body and surroundings) decreases, so each subsequent 10°C drop takes longer.

In the first interval: average excess temperature over surroundings = 65 − 30 = 35°C. In the second interval: average excess temperature = 55 − 30 = 25°C.

Ratio of rates = 35:25 = 7:5. Since the temperature drop (10°C) is the same, the times are in ratio 5:7. First interval = 5 min → second interval = 7 min. ✓

This ratio method is a fast shortcut for exam problems.

Alternative Method — Using the Exponential Form

From the differential equation: $T = T_s + (T_0 - T_s)e^{-kt}$

Let $t = 0$ when $T = 70°C$ :

At $t = 5$ min: $60 = 30 + 40e^{-5k}$

$e^{-5k} = \frac{30}{40} = \frac{3}{4}$

$-5k = \ln(3/4) \Rightarrow k = -\frac{1}{5}\ln(3/4) = \frac{\ln(4/3)}{5}$

Let $t'$ be time to cool from 70°C to 50°C: $50 = 30 + 40e^{-kt'}$ $e^{-kt'} = \frac{20}{40} = \frac{1}{2}$ $-kt' = -\ln 2 \Rightarrow t' = \frac{\ln 2}{k} = \frac{5\ln 2}{\ln(4/3)}$

$t' = \frac{5 \times 0.693}{0.288} \approx 12.03$ min

Time from 60°C to 50°C = $t' - 5 = 12.03 - 5 \approx 7.03$ min ≈ 7 min ✓

Both methods agree. The approximate method is faster for exams.

Common Mistake

Students often assume the same time for both intervals because the temperature drop is the same (10°C each). This is wrong — the body is cooling in a non-linear way (exponential cooling). Each successive 10°C drop takes longer because the driving temperature difference keeps decreasing.

Another common error: using the wrong $T_s$ (surroundings temperature). Always check what temperature the surroundings are at — different problems use different ambient temperatures. Using $T_s = 0$ when the problem states $T_s = 30°C$ gives a completely wrong answer.

JEE Main asks Newton’s cooling problems almost every year. The two solution methods (approximate average temperature method vs exact exponential form) give slightly different answers for larger temperature drops. For small drops (<10–15°C compared to the excess temperature), the approximate method is acceptable and faster. For large drops or when exact values are needed, use the exponential form.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using the Exponential Form

Common Mistake

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