Newton's Law of Cooling - Formulas, Limitations, Examples

What Is Newton’s Law of Cooling?

Newton’s law of cooling describes the rate at which an exposed body changes temperature through radiation, which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small.

Definition: According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings.

Table of Contents

• Formula
• Derivation
• Limitations
• Solved Examples

Newton’s law of cooling is given by, dT/dt = k(T_t – T_s)

Where,

• T_t = Temperature of the body at time t
• T_s = Temperature of the surrounding
• k = Positive constant that depends on the area and nature of the surface of the body under consideration.

Newton’s Law of Cooling Formula

Greater the difference in temperature between the system and its surrounding, the more rapidly the heat is transferred, i.e., the more rapidly the body temperature of body changes. Newton’s law of cooling formula is expressed by,

T(t) = T_s + (T_o – T_s) e^-kt

Where,

• t = time,
• T(t) = Temperature of the given body at time t
• T_s = Surrounding temperature
• T_o = Initial temperature of the body
• k = Constant

Newton’s Law of Cooling Derivation

For a small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference and the surface area exposed.

dQ/dt ∝ (q – q_s)], where q and q_s are temperatures corresponding to the object and surroundings.

From the above expression , dQ/dt = -k[q – q_s)] . . . . . . . . (1)

This expression represents Newton’s law of cooling. It can be derived directly from Stefan’s law, which gives,

k = [4eσ×θ³_o/mc] A . . . . . (2)

Now, dθ/dt = -k[θ – θ_o]

⇒

where,

q_i = initial temperature of the object

q_f = final temperature of the object

ln (q_f – q₀)/(q_i – q₀) = kt

(q_f – q₀) = (q_i – q₀) e^-kt

q_f = q₀ + (q_i – q₀) e ^-kt . . . . . . (3).

⇒ Check: Heat transfer by conduction

Methods to Apply Newton’s Law of Cooling

Sometimes, when we need only approximate values from Newton’s law, we can assume a constant rate of cooling, which is equal to the rate of cooling corresponding to the average temperature of the body during the interval.

i.e., dθ\dt = k(<q> – q₀) . . . . . . . (4)

If q_i and q_f be the initial and final temperature of the body, then,

<q> = (q_i + q_f)/2 . . . . . (5)

Remember, equation (5) is only an approximation and equation (1) must be used for exact values.

Limitations of Newton’s Law of Cooling

• The difference in temperature between the body and surroundings must be small
• The loss of heat from the body should be by radiation only
• The major limitation of Newton’s law of cooling is that the temperature of the surroundings must remain constant during the cooling of the body

Solved Examples

Example 1: A body at a temperature of 40ºC is kept in a surrounding of constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.

Solution:

From Newton’s law of cooling, q_f = q_i e^-kt

Now, for the interval in which temperature falls from 40 to 35^oC.

(35 – 20) = (40 – 20) e^-k.10

e^-10k = 3/4

k = [ln 4/3]/10 . . . . (a)

Now, for the next interval,

(30 – 20) = (35 – 20)e^-kt

e^-kt = 2/3

kt = ln 3/2 . . . . (b)

From equations (a) and (b),

t = 10 × [ln(3/2)/ln(4/3)]= 14.096 min.

Aliter : (by approximate method)

For the interval in which temperature falls from 40 to 35^oC

<q> = (40 + 35)/2 = 37.5ºC

From equation (4),

dθ/dt = k(<q> – q₀)

(35 – 40)/10 = k(37.5 – 20)

k = 1/32 min^-1

Now, for the interval in which temperature falls from 35^oC to 30^oC

<q> = (35 + 30)/2 = 32.5^oC

From equation (4),

(30 – 35)/t = (32.5 – 20)

Therefore, the required time t = 5/12.5 × 35 = 14 min.

Example 2: The oil is heated to 70^oC. It cools to 50^oC after 6 minutes. Calculate the time taken by the oil to cool from 50^oC to 40^oC given the surrounding temperature T_s = 25^oC.

Solution:

Given:

The temperature of oil after 6 min, T(t) = 50^oC,

• T_s = 25^oC,
• T_o = 70^oC,
• t = 6 min

On substituting the given data in Newton’s law of cooling formula, we get,

T(t) = T_s + (T_s – T_o) e^-kt

-kt ln = [ln T(t) – T_s]/T_o – T_s

-kt = [ln 50 – 25]/70 – 25 = ln 0.555

k = – (-0.555/6) = 0.092

If T(t) = 45^oC (average temperature as the temperature decreases from 50^oC to 40^oC)

Time taken is -kt ln e = [ln T(t) – T_s]/[T_o – T_s]

-(0.092) t = ln 45 – 25/[70 – 25]

-0.092 t = -0.597

t = -0.597/-0.092 = 6.489 min.

Example 3: Water is heated to 80^oC for 10 min. How much would be the temperature if k = 0.56 per min and the surrounding temperature is 25^oC?

Solution:

Given:

• T_s = 25^oC,
• T_o = 80^oC,
• t = 10 min,
• k = 0.56

Now, substituting the above data in Newton’s law of cooling formula,

T(t) = T_s + (T_o – T_s) × e^-kt

= 25 + (80 – 25) × e^-0.56 = 25 + [55 × 0.57] = 56.35 ^oC

The temperature cools down from 80^oC to 56.35 ^oC after 10 min.

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