Newton’s Law of Cooling

Q: What is the formula for temperature in Newton's law of cooling?

The temperature at time t is θ = θ₀ + (θi - θ₀)e^-kt, where θ₀ is the temperature of the surroundings and θi the initial temperature. The excess over the surroundings decays exponentially, so the body approaches room temperature ever more slowly.

Q: How is Newton's law of cooling derived from Stefan's law?

The net radiation loss is eσA(T⁴ - T₀⁴). For a small difference ΔT, T⁴ - T₀⁴ ≈ 4T₀³ ΔT, so the loss becomes proportional to the temperature difference, which is Newton's law.

Q: What does the cooling constant depend on?

The cooling constant is k = 4eσAT₀³/ms. It increases with surface area and emissivity and decreases with the heat capacity of the body. For spheres of the same material it is inversely proportional to the radius.

Q: What is the average temperature method in Newton's law of cooling?

For a small fall from θ₁ to θ₂ in time t, the average rate (θ₁ - θ₂)/t is set equal to k((θ₁ + θ₂)/2 - θ₀). It avoids logarithms and gives answers close to the exact exponential solution when the drop is small.

PhysicsThermal Properties Of MatterFor NEET aspirants

Newton's law of cooling states that the rate at which a body cools is proportional to the difference between its temperature and that of its surroundings, provided this difference is small: $- \frac{d θ}{d t} = k (θ - θ_{0})$ . Its solution is an exponential decay of the temperature excess. This page derives Newton's law of cooling from Stefan's law, solves it exactly and by the quick average-temperature method, shows its graphs and limits, and covers the problem types seen in NEET and JEE Main.

On this page1Statement2Derivation from Stefan's law3Exact solution4Graphs5Average method6Limitations7Verification8Comparing specific heats

Key Formulas - Quick Reference

★ Must learnNewton's law: $- \frac{d θ}{d t} = k (θ - θ_{0})$ (small $θ - θ_{0}$ , constant $θ_{0}$ )
★ Must learnExact solution: $θ = θ_{0} + (θ_{i} - θ_{0}) e^{- k t}$ ; time: $t = \frac{1}{k} ln \frac{θ _{1} - θ _{0}}{θ _{2} - θ _{0}}$
★ Must learnAverage method: $\frac{θ _{1} - θ _{2}}{t} = k (\frac{θ _{1} + θ _{2}}{2} - θ_{0})$
From Stefan's law: $k = \frac{4 e σ A T _{0}^{3}}{m s}$ (radiation only)
Rate of heat loss: $- \frac{d Q}{d t} = m s k (θ - θ_{0})$
Equal times: the excess $(θ - θ_{0})$ is multiplied by the same factor $e^{- k t}$
Same surface, same excess: $\frac{m _{1} s _{1} + W}{t _{1}} = \frac{m _{2} s _{2} + W}{t _{2}}$ for the same fall in temperature

1. Statement of Newton's Law of Cooling

★ Must learn

The rate of loss of heat of a body is directly proportional to the difference in temperature between the body and its surroundings, provided the difference is small:

- \frac{d Q}{d t} \propto (θ - θ_{0}) \Rightarrow - \frac{d θ}{d t} = k (θ - θ_{0})

$θ$ = temperature of the body, $θ_{0}$ = constant temperature of the surroundings, $k$ = cooling constant (unit $s^{- 1}$ or $min^{- 1}$ ). Because only a difference appears, $θ$ may be in $^{\circ} C$ or K.

Figure 1: Newton's law of cooling: for a small temperature excess, the rate of cooling is proportional to

(θ - θ_{0})

. The constant

k

depends on the surface area, the nature of the surface and the heat capacity

m s

of the body.

2. Derivation from the Stefan-Boltzmann Law

A body (area $A$ , emissivity $e$ , mass $m$ , specific heat $s$ ) at $T$ in surroundings at $T_{0}$ loses heat at the net rate $- \frac{d Q}{d t} = e σ A (T^{4} - T_{0}^{4})$ .
Write $T = T_{0} + Δ T$ with $Δ T ≪ T_{0}$ : $T^{4} = T_{0}^{4} (1 + \frac{Δ T}{T _{0}})^{4} \approx T_{0}^{4} (1 + \frac{4Δ T}{T _{0}})$ .
So $T^{4} - T_{0}^{4} \approx 4 T_{0}^{3} Δ T$ and $- \frac{d Q}{d t} = 4 e σ A T_{0}^{3} Δ T$ .
With $- \frac{d Q}{d t} = m s (- \frac{d T}{d t})$ :
$- \frac{d T}{d t} = \frac{4 e σ A T _{0}^{3}}{m s} (T - T_{0}) = k (T - T_{0})$

Figure 2: For surroundings at

300 K

, Newton's straight line is the tangent to the exact Stefan curve at the origin. At an excess of

30 K

they already differ by about

14%

: Newton's law is reliable only for small temperature differences.

Key idea

Newton's law is the small-difference (linear) form of Stefan's law:

T^{4} - T_{0}^{4} \approx 4 T_{0}^{3} (T - T_{0})

The law also describes cooling by convection in a steady draught (forced convection), which is how Newton first found it. What $k$ depends on: the area $A$ and nature of the surface ( $e$ ) increase it; a larger heat capacity $m s$ decreases it. For spheres of the same material, $k \propto \frac{A}{m} \propto \frac{1}{r}$ .

3. Exact Solution: Exponential Cooling

Separate the variables and integrate from $θ_{i}$ at $t = 0$ to $θ$ at time $t$ :

\int_{θ_{i}}^{θ} \frac{d θ}{θ - θ _{0}} = - k \int_{0}^{t} d t \Rightarrow ln \frac{θ - θ _{0}}{θ _{i} - θ _{0}} = - k t

★ Must learn

θ = θ_{0} + (θ_{i} - θ_{0}) e^{- k t}

The temperature excess decays exponentially. At $t = 0$ , $θ = θ_{i}$ ; as $t \to \infty$ , $θ \to θ_{0}$ . The time to cool from $θ_{1}$ to $θ_{2}$ is $t = \frac{1}{k} ln \frac{θ _{1} - θ _{0}}{θ _{2} - θ _{0}}$ .

Figure 3: Solved Example 1 drawn exactly:

θ = θ_{0} + (θ_{i} - θ_{0}) e^{- k t}

. The excess falls by the same factor in equal times, so the cooling slows down as the body approaches

θ_{0}

Figure 4: Two straight-line tests of Newton's law.

ln (θ - θ_{0}) = ln (θ_{i} - θ_{0}) - k t

(left) and

- \frac{d θ}{d t} = k (θ - θ_{0})

(right). Exams often ask which graph is linear.

Exam Trick

Equal times, equal ratios. In equal time intervals the excess is multiplied by the same factor. If a body cools from $6 0^{\circ} C$ to $4 0^{\circ} C$ in 10 min in a $2 0^{\circ} C$ room (excess $40 \to 20$ , halved), in the next 10 min the excess halves again to $10$ : the body reaches $3 0^{\circ} C$ . The excess behaves like a radioactive sample with "half-life" $\frac{ln 2}{k}$ .

4. The Average-Temperature Method

For a small drop from $θ_{1}$ to $θ_{2}$ in time $t$ , replace the instantaneous rate by the average rate and the instantaneous temperature by the average temperature:

\frac{θ _{1} - θ _{2}}{t} = k (\frac{θ _{1} + θ _{2}}{2} - θ_{0})

Use the first interval to find $k$ , then apply it to the second. It needs no logarithms and is accurate when the drop is small compared with the excess; for larger drops use the exact form.

Figure 5: The average method replaces the chord slope

\frac{θ _{1} - θ _{2}}{t}

by the rate at the average temperature

\frac{θ _{1} + θ _{2}}{2}

. The two lines are nearly parallel, so the method is accurate when

θ_{1} - θ_{2}

is small compared with the excess.

Exact method

$t = \frac{1}{k} ln \frac{θ _{1} - θ _{0}}{θ _{2} - θ _{0}}$ . Always correct within the law. Needs logarithms.

Average method

$\frac{θ _{1} - θ _{2}}{t} = k (θ_{avg} - θ_{0})$ . Quick, no logarithms. Slight error when the drop is large.

Quick Recall: tap to check

\frac{d θ}{d t}

constant during cooling?

No. It is proportional to the excess, so cooling slows down as the body approaches the surroundings.

Which graph of Newton's cooling is a straight line?

ln (θ - θ_{0})

against

t

(slope

- k

), or

- \frac{d θ}{d t}

against

(θ - θ_{0})

Why can

θ

be in °C in Newton's law but not in Stefan's law?

Newton's law uses only a temperature difference; Stefan's law uses

T^{4}

, which needs kelvin.

5. Limitations and Experimental Verification

The temperature difference between the body and the surroundings must be small (roughly up to 30-40 K); otherwise use $e σ A (T^{4} - T_{0}^{4})$ .
The surroundings must stay at a constant temperature.
Heat must be lost by radiation or steady (forced) convection, not by conduction to supports; the mode of heat loss should not change during cooling.
The body's temperature should be uniform (good conductor, well stirred), and $m$ and $s$ constant (no change of state).

Verification. Hot water in a calorimeter with a stirrer and thermometer is placed in a double-walled enclosure with water between the walls (constant $θ_{0}$ ). Readings of $θ$ every minute give a straight line when $ln (θ - θ_{0})$ is plotted against $t$ .

Figure 6: Verifying Newton's law. The double-walled enclosure keeps

θ_{0}

constant; plotting

ln (θ - θ_{0})

against

t

from the readings gives a straight line.

5.1 Comparing specific heats by cooling

Two identical calorimeters with the same surface, at the same temperature, lose heat at the same rate. Fill them with equal volumes (or masses) of water and another liquid and time the same fall in temperature. Then $\frac{( m _{1} s _{1} + W ) Δ θ}{t _{1}} = \frac{( m _{2} s _{2} + W ) Δ θ}{t _{2}}$ , which gives the unknown specific heat. The liquid with the smaller heat capacity cools faster.

Figure 7: Identical calorimeters at the same temperature lose heat at the same rate (same surface, same excess). The contents with the smaller heat capacity cool faster, so timing the same temperature drop compares specific heats:

\frac{( m _{1} s _{1} + W )}{t _{1}} = \frac{( m _{2} s _{2} + W )}{t _{2}}

JEE Advanced

Heated body losing heat by Newton's law: if a heater of power $P$ warms a body that loses heat at $m s k (θ - θ_{0})$ , then $m s \frac{d θ}{d t} = P - m s k (θ - θ_{0})$ , giving $θ = θ_{0} + \frac{P}{m s k} (1 - e^{- k t})$ : a steady temperature $θ_{0} + \frac{P}{m s k}$ is reached. Surroundings not constant (a body cooling in a small closed room) or large temperature differences need the full Stefan equation, $\int \frac{d T}{T ^{4} - T _{0}^{4}}$ , solved with partial fractions.

Figure 8: Choose the method. Both give nearly the same answer when the drop is small; NEET options usually match the average method, JEE numericals the exact one.

Figure 9: Mind map of this concept for quick revision.

6. Solved Examples

Solved Example 1

A body at

4 0^{\circ} C

is kept in surroundings at a constant

2 0^{\circ} C

. Its temperature falls to

3 5^{\circ} C

10 min

. How much more time will it take to reach

3 0^{\circ} C

? Solve exactly and by the average method.

Solution:

Exact. $θ - θ_{0} = (θ_{i} - θ_{0}) e^{- k t}$ . First interval: $15 = 20 e^{- 10 k} \Rightarrow e^{- 10 k} = \frac{3}{4}$ , $k = \frac{ln ( 4/3 )}{10} min^{- 1}$ .

Second interval: $10 = 15 e^{- k t} \Rightarrow k t = ln \frac{3}{2}$ , so $t = 10 \frac{ln ( 3/2 )}{ln ( 4/3 )} = 14.1 min$ .

Average method. $\frac{40 - 35}{10} = k (37.5 - 20) \Rightarrow k = \frac{1}{35} min^{- 1}$ . Then $\frac{35 - 30}{t} = \frac{1}{35} (32.5 - 20) \Rightarrow t = \frac{5 \times 35}{12.5}$ .

Answer: about $14.1 min$ (exact); $14 min$ (average method).

Solved Example 2

A body cools from

8 0^{\circ} C

6 0^{\circ} C

5 min

in a room at

2 0^{\circ} C

. How long will it take to cool from

6 0^{\circ} C

4 0^{\circ} C

Solution:

Exact: $e^{- 5 k} = \frac{40}{60} = \frac{2}{3}$ ; second interval $e^{- k t} = \frac{20}{40} = \frac{1}{2}$ . So $t = 5 \frac{ln 2}{ln 1.5} = 8.55 min$ .

Average method: $\frac{20}{5} = k (70 - 20) \Rightarrow k = 0.08$ ; $\frac{20}{t} = 0.08 (50 - 20) \Rightarrow t = 8.33 min$ .

Answer: about $8.5 min$ (the average method gives $8.3 min$ : the drop here is not small, so the exact value is better).

Solved Example 3

A body at

9 0^{\circ} C

is placed in a room at

3 0^{\circ} C

. Its cooling constant is

k = 0.05 min^{- 1}

. Find its temperature after

10 min

Solution:

$θ = 30 + (90 - 30) e^{- 0.05 \times 10} = 30 + 60 e^{- 0.5} = 30 + 60 (0.607)$ .

Answer: $θ \approx 66. 4^{\circ} C$ .

Solved Example 4

A cup of tea cools from

9 0^{\circ} C

8 0^{\circ} C

2 min

in a room at

2 0^{\circ} C

. How long does it take to cool from

8 0^{\circ} C

7 0^{\circ} C

Solution:

$e^{- 2 k} = \frac{60}{70} = \frac{6}{7}$ ; for the next drop $e^{- k t} = \frac{50}{60} = \frac{5}{6}$ .

$t = 2 \frac{ln ( 6/5 )}{ln ( 7/6 )} = 2 \times \frac{0.1823}{0.1542}$ .

Answer: $t \approx 2.37 min$ : the same $1 0^{\circ} C$ drop takes longer as the tea gets cooler.

Solved Example 5

A body cools from

6 0^{\circ} C

4 0^{\circ} C

10 min

in a room at

2 0^{\circ} C

. What will its temperature be after the next

10 min

Solution:

The excess went from $40$ to $20$ : it halved in $10 min$ . In the next $10 min$ it halves again, to $10$ .

Answer: $θ = 20 + 10 = 3 0^{\circ} C$ .

Solved Example 6

Two identical calorimeters (water equivalent

10 g

each) hold

100 g

of water and

100 g

of a liquid. Both cool from

5 0^{\circ} C

4 0^{\circ} C

under the same conditions, taking

5 min

and

2 min

respectively. Find the specific heat of the liquid.

Solution:

Same surface and same temperatures, so the same rate of heat loss: $\frac{( 100 \times 1 + 10 ) ( 10 )}{5} = \frac{( 100 s + 10 ) ( 10 )}{2}$ .

$44 = 100 s + 10$ .

Answer: $s = 0.34 cal g^{- 1}^{\circ} C^{- 1}$ .

Solved Example 7

Two solid spheres of the same material and surface finish, radii

r

and

2 r

, are heated to the same temperature and left in the same room. Compare their initial rates of cooling.

Solution:

$k = \frac{4 e σ A T _{0}^{3}}{m s} \propto \frac{A}{m} = \frac{4 π r ^{2}}{\frac{4}{3} π r ^{3} ρ} \propto \frac{1}{r}$ .

Answer: $2 : 1$ (the smaller sphere cools twice as fast).

Solved Example 8

For a body obeying Newton's law of cooling, which graph is a straight line?
(A)

θ

against

t

(B)

ln θ

against

t

(C)

ln (θ - θ_{0})

against

t

(D)

θ

against

ln t

Solution:

Answer: (C). $ln (θ - θ_{0}) = ln (θ_{i} - θ_{0}) - k t$ is linear in $t$ with slope $- k$ . $θ$ against $t$ is an exponential curve.

Solved Example 9

Newton's law of cooling is a special case of
(A) Wien's law
(B) Kirchhoff's law
(C) Stefan's law
(D) Planck's law. Under what condition does it hold?

Solution:

Answer: (C), valid when the temperature difference between the body and its surroundings is small, so that $T^{4} - T_{0}^{4} \approx 4 T_{0}^{3} (T - T_{0})$ .

Practice Questions

A liquid cools from $7 0^{\circ} C$ to $6 0^{\circ} C$ in $5 min$ in a room at $3 0^{\circ} C$ . Using the average method, find the time to cool from $6 0^{\circ} C$ to $5 0^{\circ} C$ .Answer: $7 min$
A body's excess temperature falls from $32 K$ to $16 K$ in $6 min$ . Find the excess after a further $12 min$ .Answer: $4 K$
Find $k$ for a body whose excess falls to $\frac{1}{e}$ of its value in $20 min$ .Answer: $0.05 min^{- 1}$
Hot water cools from $6 0^{\circ} C$ to $5 0^{\circ} C$ in $10 min$ in a room at $1 0^{\circ} C$ . What is its temperature after the next $10 min$ (exact)?Answer: $4 2^{\circ} C$ (excess $50 \to 40 \to 32$ )
State two conditions under which Newton's law of cooling holds.Answer: Small temperature difference; constant surroundings (also: no change in the mode of heat loss)
Why does a thin wire cool faster than a thick rod of the same metal at the same temperature?Answer: $k \propto \frac{A}{m}$ is larger for the thin wire

Common Mistakes to Avoid

Watch out

Applying Newton's law to large temperature differences (for example a body at $50 0^{\circ} C$ ); use Stefan's law there.
Assuming the rate of cooling is constant and dividing total drop by total time over a long interval.
Using the average method for a large drop and expecting the exact answer.
Using $θ$ instead of the excess $(θ - θ_{0})$ in the exponential: it is the excess that decays.
Saying the cooling constant is the same for all bodies. It depends on $A$ , $e$ , $m$ and $s$ .
In calorimeter comparisons, forgetting the calorimeter's own water equivalent $W$ in both heat capacities.
Thinking a body cools to the room temperature in finite time exactly; the approach is exponential.

Frequently Asked Questions

What is Newton's law of cooling?

Newton's law of cooling says that the rate at which a body loses heat is proportional to the difference between its temperature and that of its surroundings, provided the difference is small. In symbols, $- \frac{d θ}{d t} = k (θ - θ_{0})$ .

What is the formula for temperature in Newton's law of cooling?

The temperature at time $t$ is $θ = θ_{0} + (θ_{i} - θ_{0}) e^{- k t}$ , where $θ_{0}$ is the temperature of the surroundings and $θ_{i}$ the initial temperature. The excess over the surroundings decays exponentially, so the body approaches room temperature ever more slowly.

How is Newton's law of cooling derived from Stefan's law?

The net radiation loss is $e σ A (T^{4} - T_{0}^{4})$ . For a small difference $Δ T$ , $T^{4} - T_{0}^{4} \approx 4 T_{0}^{3} Δ T$ , so the loss becomes proportional to the temperature difference, which is Newton's law.

What are the limitations of Newton's law of cooling?

It holds only when the temperature difference is small, the surroundings stay at a constant temperature, heat is lost mainly by radiation or steady convection, and the body's temperature is uniform without any change of state.

What does the cooling constant depend on?

The cooling constant is $k = \frac{4 e σ A T _{0}^{3}}{m s}$ . It increases with surface area and emissivity and decreases with the heat capacity of the body. For spheres of the same material it is inversely proportional to the radius.

What is the average temperature method in Newton's law of cooling?

For a small fall from $θ_{1}$ to $θ_{2}$ in time $t$ , the average rate $\frac{θ _{1} - θ _{2}}{t}$ is set equal to $k (\frac{θ _{1} + θ _{2}}{2} - θ_{0})$ . It avoids logarithms and gives answers close to the exact exponential solution when the drop is small.

How is Newton's law of cooling asked in NEET?

NEET usually gives a body cooling between two temperatures in a known time and asks the time for the next drop, solved by the average method, or asks which graph is linear and what the cooling rate depends on.

What Newton's law of cooling problems come in JEE Main?

JEE Main uses the exact exponential solution with logarithms, comparison of specific heats from cooling times, spheres of different sizes, graphs of log excess against time, and the link between Newton's law and Stefan's law.

Previous year questions on Newton’s Law of Cooling

3 questions from past papers, each with a step-by-step solution.

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