Q.A pan filled with hot food cools from 94 ∘C to 86 ∘C in 2 minutes when the room temperature is at 20 ∘C. How long will it take to cool from 71 ∘C to 69 ∘C?
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 Start your 14-day free trial to unlock the full solution →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 Ts is the constant temperature of the surroundings (the "ambient" temperature), then:
dtdT∝−(T−Ts)
The negative sign is crucial: it tells us the temperature decreases when T>Ts (cooling) and increases when T<Ts (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:
dtdT=−k(T−Ts)
dtdT=−k(T−Ts)
This is a simple first-order differential equation. Its solution, which gives the temperature at any time, is:
T(t)=Ts+(T0−Ts)e−kt
where T0 is the initial temperature of the object at t=0.
What the Solution Tells You
- Exponential decay of the temperature difference. The quantity (T−Ts) shrinks exponentially toward zero. The object never exactly reaches Ts 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 Ts is the asymptote. The object's temperature approaches Ts from above (cooling) or below (heating).
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 Ts 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 T0=90, Ts=20, t=5 min, T(5)=60.
From the solution: 60=20+(90−20)e−5k → 40=70e−5k → e−5k=74 → k=−51ln(74)≈0.112 per minute.
Step 2: Find T(10): T(10)=20+70e−10k=20+70(e−5k)2=20+70(74)2=20+70⋅4916=20+491120≈42.86∘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 …
Using the average-temperature form of Newton's Law of Cooling, tT1−T2=k(2T1+T2−T0), with room temperature 20∘C.
First interval (94→86∘C in 2 min): average excess =90−20=70∘C; rate =4∘C/min ⇒k=4/70=2/35 min−1. …
Using the standard average-temperature form of Newton's Law of Cooling, the cooling constant from the first interval gives a time of 0.7 minutes (42 seconds) for the second interval.
Newton's Law of Cooling says the rate of heat loss is proportional to the excess temperature over the surroundings. For a temperature drop over a short interval, we can use the practical (average-temperature) form:
tT1−T2=k(2T1+T2−T0),
where T0=20∘C is the room temperature.
Step 1 - Find k from the first interval (94∘C to 86∘C in 2 minutes)
Average temperature: 294+86=90∘C. Excess over the room: 90−20=70∘C.
Rate of cooling: 294−86=4 ∘C/min.
4=k×70⇒k=704=352 min−1.
Step 2 - Apply k to the second interval (71∘C to 69∘C)
Average temperature: 271+69=70∘C. Excess over the room: 70−20=50∘C.
Rate of cooling now: k×50=352×50=35100=720 ∘C/min. …
There's a shortcut that never needs a numeric value of k: since k is the same constant in both intervals, cooling rate is simply proportional to excess temperature over the room, so t1t2=ΔT1ΔT2×Tˉ2−T0Tˉ1−T0. Plugging in ΔT1=8∘C over t1=2 min at average excess 70∘C, and ΔT2=2∘C at average excess 50∘C, gives t2=2×82×5070=0.7 min directly, by pure proportion. The physical insight worth keeping: naively scaling the f …
- COMEDK 2024Set 2024-M1 markMCQQ.A glass of hot water cools from 90∘C to 70∘C in 3 minutes when the temperature of surroundings is 20∘C. What is the time taken by the glass of hot water to cool from 60∘C to 40∘C if the surrounding temperature remains the same at 20∘C ? (A) 15 minutes (B) 6 minutes (C) 12 minutes (D) 10 minutes
›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:
dtdT=−k(T−Ts)
which integrates to
ln(T0−TsT−Ts)=−kt
where T0 is initial temperature, Ts is surrounding temperature, and k is a positive constant.
Step-by-step solution:
- Set up the first cooling interval. Initial temperature T1=90∘C, final temperature T2=70∘C, surrounding Ts=20∘C, time t=3 minutes. Using the integrated form:
ln(90−2070−20)=−k⋅3
ln(7050)=−3k
ln(75)=−3k
So
k=−31ln(75)=31ln(57)
- Now consider the second cooling interval. Initial temperature T1′=60∘C, final temperature T2′=40∘C, same Ts=20∘C. Let the unknown time be t′. Apply the same formula:
ln(60−2040−20)=−kt′
ln(4020)=−kt′
ln(21)=−kt′
So
kt′=ln2
- Substitute the value of k from step 1.
(31ln57)t′=ln2
t′=3⋅ln(7/5)ln2
- Evaluate numerically. ln2≈0.6931, ln(7/5)=ln1.4≈0.3365.
t′≈3×0.33650.6931≈3×2.059≈6.18 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:
ln(7/5)ln2=log1.42 …
- KCET 2020Set A-11 markMCQQ.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 ⇒ same density ρ. Same mass m. Therefore
V=ρmis 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:
dtdQ∝A(T−T0)
The rate of temperature fall is then
dtdT=mc1dtdQ∝mA
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→0 has A→∞ for fixed volume, since A≈2V/t).
So, at equal volume:
Asphere<Acube<Athin plate
Numerical illustration for V=1 unit:
- Sphere: r=(4π3)1/3=0.620⇒A=4πr2≈4.84 …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.