Solved Examples Newton’s Law of Cooling

Example 1: 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?

Solution:

Average of 94 °C and 86 °C is 90 °C, 

• T₂ = 90 °C
• T₁ = 20 °C

Drop in tem. of food is 8 °C in 2 minutes.

According to Newton’s law of cooling,

– dQ/dt = k(T₂ –T₁) 

 8 °C /2 min = k(90 – 20)

4 = k(70)………(1)

Average of 69 °C and 71 °C is 70 °C

• T₂ = 70 °C
• T₁ = 20 °C

According to Newton’s law of cooling,

2 °C /dt = k(70 – 20) ……(2)        

From equation (1) and (2),

Change in time = 0.7 min = =42 sec

Thus, the food will take 42 sec to cool from 71 °C to 69 °C.

Example 2: 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:

Given,

• q_i = (40 – 20)ºC
• q_f = (35 – 20)ºC

According to Newtons law of cooling

 q_f  = q_i e^-kt

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

(35 – 20) = (40 – 20) e^-(10k)

e^-10k = 3/4

-10k = (ln 4/3)

k = 0.2876/10

k = 0.02876

Now using Newon’s Formula again,

(30 – 20) = (35 – 20)e^-kt

10 = 15e^-kt

e^-kt = 2/3

-kt = ln(2/3)

t = 0.40546/k

Using the value of the k,

t = 0.40546/0.02876

t = 14.098 min

Thus, the time taken by body to reach the temp of 30ºC is 14.098 min

Example 3: 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,

Temperature of oil after 6 min i.e. T(t) is equal to 50 ºC

• Ambient Temperature T_s = 25 ºC
• Temperature of Oil, T_o = 70 ºC
• Time to Cool to 50ºC = 6 min

According to Newton’s law of cooling,

T(t) = T_s + (T₀ – T_s) e^-kt

{T(t) – T_s}/(T_o – T_s) = e^-kt

-kt = ln[(T(t) – T_s)/(T_o – T_s)] ………(1)        

Substitute the values

-kt = ln[(50 – 25)/(70 – 25)] 

-k = (ln 0.55556)/6

k = 0.09796

Average Temperature from 50 ºC to 40 ºC is equal to 45 ºC 

Againg using Newton’s Law of cooling

-(0.09796)t = ln[(45 – 25)/(70 – 25)]

-0.09796t = ln(0.44444)

0.09796t = 0.81093

t = 0.09796/0.58778 = 8.278 min

Thus, the time take by oil to cool from 50 ºC to 40 ºC is 8.278 min

Example 4: Water is heated to 80 ºC for 10 min. How much would be its temperature in degrees Celsius, if k = 0.056 per min and the surrounding temperature is 25 ºC?

Solution:

Given,

• Ambient Temperature T_s = 25 ºC
• Temperature of water T₀ = 80 ºC
• Time for which Water is heated (t) = 10 min
• Value of constant k = 0.056.

According to Newton’s law of cooling,

T(t) = T_s + (T₀ – T_s) e^-kt

Substituting the value

T(t)= 25 + (80 – 25)e^-(0.056×10) 

T(t) = 25 + 55 e^-(0.056×10)

T(t) = 25 + 31.42

T(t) = 56.42

After 10 min the temperature of water would be 56.42 ºC.

Newton’s Law of Cooling is the fundamental law that describes the rate of heat transfer by a body to its surrounding through radiation. This law state that the rate at which the body radiate heats is directly proportional to the difference in the temperature of the body from its surrounding, given that the difference in temperature is low. i.e. the higher the difference between the temperature of the body and its surrounding the more heat is lost and the lower the temperature the less heat is lost. Newton’s Law of Cooling is a special case of Stefan-Boltzmann’s Law.

In this article, we will learn about, Newton’s Law of Cooling, Newton’s Law of Cooling Formula, its Derivation, Examples, and others in detail.