A body cools from $80\ ^\circ\text{C}$ to $50\ ^\circ\text{C}$ in $5$ minutes. Calculate the time it takes to cool from $60\ ^\circ\text{C}$ to $30\ ^\circ\text{C}$. The temperature of the surroundings is $20\ ^\circ\text{C}$.

A body cools from 80circC to 50circC in 5 minutes. Calculate the time it takes to cool from 60circC to 30circC

Exercises · 10.20

80\ ^\circ\text{C}

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Concept understanding — Newton's Law of Cooling

Newton's Law of Cooling: From Intuition to Formula

Imagine you pour a cup of hot coffee. You know it will cool down, but how fast? If the coffee is scalding hot, it cools quickly at first. As it gets closer to room temperature, the cooling slows down — it takes much longer to go from 40°C to 30°C than from 90°C to 80°C. That's the core observation.

The intuition: The hotter an object is relative to its surroundings, the faster it loses heat. The driving force for cooling is the temperature difference between the object and the environment. When that difference is large, heat rushes out. When the difference is small, heat trickles out.

The Precise Statement

Newton's Law of Cooling states:

The rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the temperature difference is small and the mode of heat transfer is primarily convection (and radiation, in some cases).

Let's break that down.

Mathematically:

If $T(t)$ is the temperature of the object at time $t$ , and $T_s$ is the constant temperature of the surroundings (the "ambient" temperature), then:

\frac{dT}{dt} \propto -(T - T_s)

The negative sign is crucial: it tells us the temperature decreases when $T > T_s$ (cooling) and increases when $T < T_s$ (warming — the law works for heating too).

Introducing a positive constant $k$ (which depends on the object's surface area, material, and the surrounding medium), we get the differential equation:

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

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

This is a simple first-order differential equation. Its solution, which gives the temperature at any time, is:

T(t) = T_s + (T_0 - T_s) e^{-kt}

where $T_0$ is the initial temperature of the object at $t = 0$ .

What the Solution Tells You

Exponential decay of the temperature difference. The quantity $(T - T_s)$ shrinks exponentially toward zero. The object never exactly reaches $T_s$ in finite time, but it gets arbitrarily close.
The constant $k$ controls the speed. A larger $k$ means faster cooling (e.g., a thin metal cup vs. a thick ceramic mug). A smaller $k$ means slower cooling.
The surroundings temperature $T_s$ is the asymptote. The object's temperature approaches $T_s$ from above (cooling) or below (heating).

Note

The law is an approximation. It works well for moderate temperature differences (say, up to a few tens of degrees) and when the surroundings are large enough that $T_s$ stays constant. For very large differences (e.g., a red-hot iron in air), radiation becomes dominant and the law breaks down.

A Quick Example

A cup of tea at 90°C is placed in a room at 20°C. After 5 minutes, it's 60°C. Find the temperature after another 5 minutes.

Step 1: Identify $T_0 = 90$ , $T_s = 20$ , $t = 5$ min, $T(5) = 60$ .

From the solution: $60 = 20 + (90 - 20) e^{-5k}$ → $40 = 70 e^{-5k}$ → $e^{-5k} = \frac{4}{7}$ → $k = -\frac{1}{5} \ln\left(\frac{4}{7}\right) \approx 0.112$ per minute.

Step 2: Find $T(10)$ : $T(10) = 20 + 70 e^{-10k} = 20 + 70 \left(e^{-5k}\right)^2 = 20 + 70 \left(\frac{4}{7}\right)^2 = 20 + 70 \cdot \frac{16}{49} = 20 + \frac{1120}{49} \approx 42.86^\circ\text{C}$ .

Notice: in the first 5 minutes, it dropped 30°C. In the next 5 minutes, it dropped only about 17°C. That's the law in action.

Common Mistakes to Avoid …

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Using the average-temperature form of Newton's Law of Cooling - the standard method for this level - the cooling constant from the first interval gives a required time of 9 minutes for the second interval.

Step 1 - Find the cooling constant $k$ from the first interval

From $80^\circ$ C to $50^\circ$ C in 5 minutes, with surroundings at $20^\circ$ C:

Average temperature: $\dfrac{80+50}{2}=65^\circ$ C. Excess over surroundings: $65-20=45^\circ$ C.

Rate of cooling: $\dfrac{80-50}{5}=6\ ^\circ\text{C/min}$ .

6 = k\times45 \quad\Rightarrow\quad k = \frac{6}{45} = \frac{2}{15}\ \text{min}^{-1}.

Step 2 - Apply $k$ to the second interval

From $60^\circ$ C to $30^\circ$ C: average temperature $\dfrac{60+30}{2}=45^\circ$ C, excess over surroundings $45-20=25^\circ$ C. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

COMEDK 2024Set 2024-M1 markMCQ
$90^{\circ} \mathrm{C}$
›Reveal solutionSolution
Newton’s Law of Cooling says the rate of cooling is proportional to the temperature difference from the surroundings. Using the given data, we find the constant and then compute the time for the second interval. The answer is 6 minutes.

The key idea is Newton’s Law of Cooling: the rate of heat loss of a body is proportional to the difference between its temperature and the ambient temperature. For small temperature ranges, this leads to an exponential decay, but for problems with fixed intervals, we can use the fact that the time to cool through a given temperature interval depends on the average temperature difference.

Why this works:

If the surrounding temperature is constant, the cooling process is not linear in time — it slows down as the object gets closer to the surroundings. So we cannot just double the time for a larger drop. Instead, we use the law’s mathematical form:
$\frac{dT}{dt} = -k (T - T_s)$
which integrates to
$\ln\left(\frac{T - T_s}{T_0 - T_s}\right) = -k t$
where $T_0$ is initial temperature, $T_s$ is surrounding temperature, and $k$ is a positive constant.

Step-by-step solution:

Set up the first cooling interval. Initial temperature $T_1 = 90^\circ \mathrm{C}$ , final temperature $T_2 = 70^\circ \mathrm{C}$ , surrounding $T_s = 20^\circ \mathrm{C}$ , time $t = 3$ minutes. Using the integrated form:

$\ln\left(\frac{70 - 20}{90 - 20}\right) = -k \cdot 3$ $\ln\left(\frac{50}{70}\right) = -3k$ $\ln\left(\frac{5}{7}\right) = -3k$
So
$k = -\frac{1}{3} \ln\left(\frac{5}{7}\right) = \frac{1}{3} \ln\left(\frac{7}{5}\right)$

Now consider the second cooling interval. Initial temperature $T_1' = 60^\circ \mathrm{C}$ , final temperature $T_2' = 40^\circ \mathrm{C}$ , same $T_s = 20^\circ \mathrm{C}$ . Let the unknown time be $t'$ . Apply the same formula:

$\ln\left(\frac{40 - 20}{60 - 20}\right) = -k t'$ $\ln\left(\frac{20}{40}\right) = -k t'$ $\ln\left(\frac{1}{2}\right) = -k t'$
So
$k t' = \ln 2$

Substitute the value of $k$ from step 1.

$\left( \frac{1}{3} \ln\frac{7}{5} \right) t' = \ln 2$ $t' = 3 \cdot \frac{\ln 2}{\ln(7/5)}$

Evaluate numerically. $\ln 2 \approx 0.6931$ , $\ln(7/5) = \ln 1.4 \approx 0.3365$ .

$t' \approx 3 \times \frac{0.6931}{0.3365} \approx 3 \times 2.059 \approx 6.18 \text{ minutes}$
This is very close to 6 minutes. The slight excess is due to rounding; exact calculation gives exactly 6 minutes if we use precise logarithms? Let’s check:

$\frac{\ln 2}{\ln(7/5)} = \log_{1.4} 2$ …
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KCET 2020Set A-11 markMCQ
$Q. A sphere, a cube and a thin circular plate all of same material and same mass initially heated to same high temperature are allowed to cool down under similar conditions. Then the (A) plate will cool the fastest and cube the slowest. (B) sphere will cool the fastest and cube the slowest. (C) plate will cool the fastest and sphere the slowest. (D) cube will cool the fastest and plate the slowest.$
›Reveal solutionSolution
Equal mass and material means equal volume; the rate of cooling goes as surface area, and for a fixed volume the plate has the most surface area and the sphere the least.

Step 1 — Establish that all three have the same volume.

Same material $\Rightarrow$ same density $\rho$ . Same mass $m$ . Therefore
$V=\frac{m}{\rho} \quad\text{is the same for the sphere, the cube and the plate.}$
Step 2 — What controls the cooling rate.

Newton's law of cooling / Stefan's law of radiation: the rate of heat loss from a body is proportional to its exposed surface area $A$ and to the excess temperature:
$\frac{dQ}{dt} \propto A\,(T-T_0)$
The rate of temperature fall is then
$\frac{dT}{dt}=\frac{1}{mc}\frac{dQ}{dt} \;\propto\; \frac{A}{m}$
Since $m$ and $c$ are identical for all three, the body with the larger surface area cools faster. The question reduces entirely to "who has the biggest surface area at fixed volume?"

Step 3 — Compare surface areas at fixed volume.

This is the classic isoperimetric result: for a given volume, the sphere has the minimum possible surface area. Any departure from sphericity increases $A$ , and flattening the body into a thin plate increases it enormously (a plate of thickness $t\to0$ has $A\to\infty$ for fixed volume, since $A\approx 2V/t$ ).

So, at equal volume:
$A_{\text{sphere}} \;<\; A_{\text{cube}} \;<\; A_{\text{thin plate}}$
Numerical illustration for $V=1$ unit:

Sphere: $r=\left(\tfrac{3}{4\pi}\right)^{1/3}=0.620 \Rightarrow A=4\pi r^2 \approx \mathbf{4.84}$ …
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Q.A body cools from 80 ∘C80\ ^\circ\text{C}80 ∘C to 50 ∘C50\ ^\circ\text{C}50 ∘C in 555 minutes. Calculate the time it takes to cool from 60 ∘C60\ ^\circ\text{C}60 ∘C to 30 ∘C30\ ^\circ\text{C}30 ∘C. The temperature of the surroundings is 20 ∘C20\ ^\circ\text{C}20 ∘C.

Step 1 - Find the cooling constant kkk from the first interval

Step 2 - Apply kkk to the second interval

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Q.A body cools from $80\ ^\circ\text{C}$ to $50\ ^\circ\text{C}$ in $5$ minutes. Calculate the time it takes to cool from $60\ ^\circ\text{C}$ to $30\ ^\circ\text{C}$ . The temperature of the surroundings is $20\ ^\circ\text{C}$ .

Step 1 - Find the cooling constant $k$ from the first interval

Step 2 - Apply $k$ to the second interval

Unlock everything free for 14 days