**Newton’s law of cooling** describes the rate at which an exposed body changes temperature. According to this law, the rate of heat loss of a body is directly proportional to the difference in temperature between the body and its surrounding environment, provided the temperature difference is not too large. This law explains why a hot cup of coffee cools faster initially when the temperature difference between the coffee and the room is greater, and slows down as the coffee’s temperature approaches room temperature. Newton’s law of cooling is widely used in various fields such as physics, engineering, and environmental science to model the cooling process of objects.

Newton’s law of cooling is mathematically expressed as:

$$\frac{dT}{dt} = -k\left ( T – T_{env} \right )$$where dT/dt is the rate of change of the temperature T of the object, T_{env} is the ambient temperature of the environment, and k is a positive constant that depends on the characteristics of the object and its surroundings.

## Practice problems

### Problem #1

A hot cup of coffee cools down from 90 ℃ to 80 ℃ in 5 minutes when placed on the table. Determine the time it takes for the coffee to cool from 60 ℃ to 50 ℃, considering that the temperature of the surroundings is 25 ℃.

**Solution**

**Case #1** When a hot coffee cools down from 90 ℃ to 80 ℃

The average temperature of 90 ℃ and 80 ℃ is 85 ℃, which is 60 ℃ above the room temperature. (i.e. a hot coffee cools up to 10 ℃ in 5 minutes)

From the above statement:

- Rate of cooling, dQ/dt = 10/5
- Average temperature, T
_{(t)}= 85 ℃ - Surrounding temperature, T
_{s}= 25 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 10/5 = k (85 – 25)
- 10/5 = k (60) –––––––– Equation (1)

Where, k = constant

**Case #2** When a hot coffee cools down from 60 ℃ to 50 ℃

The average temperature of 60 ℃ and 50 ℃ is 55 ℃, which is 30 ℃ above the room temperature. (i.e. a hot coffee cools up to 10 ℃ in ‘t’ time)

From the above statement:

- Rate of cooling, dQ/dt = 10/t
- Average temperature, T
_{(t)}= 55 ℃ - Surrounding temperature, T
_{s}= 25 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 10/t = k (55 – 25)
- 10/t = k (30) –––––––– Equation (2)

**Note:** Value of k remains the same in both cases

On dividing Equation (1) and Equation (2):

- (10/5) × (t/10) = 60/30
- t/5 = 20
- t = 100 minutes

Therefore, a hot coffee takes **100 minutes** to cool down from 60 ℃ to 50 ℃.

### Problem #2

A metal rod initially at a temperature of 90 ℃ cools down to 70 ℃ in 4 minutes. If the temperature of the surroundings is 20 ℃, calculate the time required for the rod to cool from 50 ℃ to 30 ℃.

**Solution**

**Case #1** When a metal rod cools down from 90 ℃ to 70 ℃

The average temperature of 90 ℃ and 70 ℃ is 80 ℃, which is 60 ℃ above the room temperature. (i.e. a metal rod cools up to 20 ℃ in 4 minutes)

From the above statement:

- Rate of cooling, dQ/dt = 20/4
- Average temperature, T
_{(t)}= 80 ℃ - Surrounding temperature, T
_{s}= 20 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 20/4 = k (80 – 20)
- 20/4 = k (60) –––––––– Equation (1)

Where, k = constant

**Case #2** When a metal rod cools down from 50 ℃ to 30 ℃

The average temperature of 50 ℃ and 30 ℃ is 40 ℃, which is 20 ℃ above the room temperature. (i.e. a metal rod cools up to 20 ℃ in ‘t’ time)

From the above statement:

- Rate of cooling, dQ/dt = 20/t
- Average temperature, T
_{(t)}= 40 ℃ - Surrounding temperature, T
_{s}= 20 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 20/t = k (40 – 20)
- 20/t = k (20) –––––––– Equation (2)

**Note:** Value of k remains the same in both cases

On dividing Equation (1) and Equation (2):

- (20/4) × (t/20) = 60/20
- t/4 = 3
- t = 12 minutes

Therefore, a metal rod takes **12 minutes** to cool down from 50 ℃ to 30 ℃.

### Problem #3

When a hot cup of soup is placed on a desk, it cools down from 100 ℃ to 80 ℃ in 6 minutes. If the temperature of the surroundings is 24 ℃, determine the time it takes for the soup to cool from 70 ℃ to 50 ℃.

**Solution**

**Case #1** When the soup cools down from 100 ℃ to 80 ℃

The average temperature of 100 ℃ and 80 ℃ is 90 ℃, which is 66 ℃ above the room temperature. (i.e. the soup cools up to 20 ℃ in 6 minutes)

From the above statement:

- Rate of cooling, dQ/dt = 20/6
- Average temperature, T
_{(t)}= 90 ℃ - Surrounding temperature, T
_{s}= 24 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 20/6 = k (90 – 24)
- 20/6 = k (66) –––––––– Equation (1)

Where, k = constant

**Case #2** When the soup cools down from 70 ℃ to 50 ℃

The average temperature of 70 ℃ and 50 ℃ is 60 ℃, which is 36 ℃ above the room temperature. (i.e. the soup cools up to 20 ℃ in ‘t’ time)

From the above statement:

- Rate of cooling, dQ/dt = 20/t
- Average temperature, T
_{(t)}= 60 ℃ - Surrounding temperature, T
_{s}= 24 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 20/t = k (60 – 24)
- 20/t = k (36) –––––––– Equation (2)

**Note:** Value of k remains the same in both cases

On dividing Equation (1) and Equation (2):

- (20/6) × (t/20) = 66/36
- t/6 = 66/36
- t = 66/6
- t = 11 minutes

Therefore, the soup takes **11 minutes** to cool down from 70 ℃ to 50 ℃.

### Problem #4

A large container holds oil with an initial temperature of 150 ℃. The oil cools down to 120 ℃ in 8 minutes. Calculate the time it will take for the oil to cool from 100 ℃ to 70 ℃, considering that the temperature of the surroundings is 23 ℃.

**Solution**

**Case #1** When the oil cools down from 150 ℃ to 120 ℃

The average temperature of 150 ℃ and 120 ℃ is 135 ℃, which is 112 ℃ above the room temperature. (i.e. the oil cools up to 30 ℃ in 8 minutes)

From the above statement:

- Rate of cooling, dQ/dt = 30/8
- Average temperature, T
_{(t)}= 135 ℃ - Surrounding temperature, T
_{s}= 23 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 30/8 = k (135 – 23)
- 30/8 = k (112) –––––––– Equation (1)

Where, k = constant

**Case #2** When the oil cools down from 100 ℃ to 70 ℃

The average temperature of 100 ℃ and 70 ℃ is 85 ℃, which is 62 ℃ above the room temperature. (i.e. the oil cools up to 30 ℃ in ‘t’ time)

From the above statement:

- Rate of cooling, dQ/dt = 30/t
- Average temperature, T
_{(t)}= 85 ℃ - Surrounding temperature, T
_{s}= 23 ℃

Applying the formula:

- – dQ/dt = h A (T
_{(t)}– T_{s}) - dQ/dt = – h A (T
_{(t)}– T_{s}) - dQ/dt = k (T
_{(t)}– T_{s}) - 30/t = k (85 – 23)
- 30/t = k (62) –––––––– Equation (2)

**Note:** Value of k remains the same in both cases

On dividing Equation (1) and Equation (2):

- (30/8) × (t/30) = 112/62
- t/8 = 112/62
- t = 896/62
- t = 14.45 minutes

Therefore, the oil takes **14.45 minutes** to cool down from 100 ℃ to 70 ℃.

## More topics

- Law of conservation of energy
- Newton’s law of universal gravitation
**Newton’s law of cooling**- Coulomb’s law
- Snell’s law
- Ohm’s law
- Hooke’s law

## External links

- https://socratic.org/precalculus/solving-exponential-and-logarithmic-equations/newtons-law-of-cooling
- https://web.math.ucsb.edu/~myoshi/cooling.pdf
- https://www.softschools.com/formulas/physics/newtons_law_of_cooling_formula/93/

Deep

Learnool.com was founded by Deep Rana, who is a mechanical engineer by profession and a blogger by passion. He has a good conceptual knowledge on different educational topics and he provides the same on this website. He loves to learn something new everyday and believes that the best utilization of free time is developing a new skill.