**Newton’s law of cooling equation** states that the **rate of heat loss** (- dQ/dt) by a body directly depends upon the **temperature difference** (ΔT) of a body and its surroundings. Here’s the equation of newton’s law of cooling: **– dQ/dt = h A (T**_{(t)}** – T**_{s}**)**

Let’s solve some problems based on this equation, so you’ll get a clear idea.

## Newton’s Law of Cooling Practice Problems

**Problem 1:** A hot coffee cools down from 90 °C to 80 °C in 5 minutes when it is placed on the table. Calculate the time taken by a hot coffee to cool down from 60 °C to 50 °C. The temperature of the surroundings is 25 °C.

Solution:**Case 1:** When a hot coffee cools down from 90 °C to 80 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 10/5

Average temperature, T_{(t)} = 85 °C

Surrounding temperature, T_{s} = 25 °C

Using the equation of newton’s law of cooling,

– 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 °C to 50 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 10/t

Average temperature, T_{(t)} = 55 °C

Surrounding temperature, T_{s} = 25 °C

Using the equation of newton’s law of cooling,

– 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) we get,

(10/5) × (t/10) = 60/30

t/5 = 20

t = 100 minutes

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

**Problem 2:** A hot metal rod cools down from 90 °C to 70 °C in 4 minutes. If the temperature of the surroundings is 20 °C, then how much time will it take to cool from 50 °C to 30 °C?

Solution:**Case 1:** When a hot metal rod cools down from 90 °C to 70 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 20/4

Average temperature, T_{(t)} = 80 °C

Surrounding temperature, T_{s} = 20 °C

Using the equation of newton’s law of cooling,

– 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 hot metal rod cools down from 50 °C to 30 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 20/t

Average temperature, T_{(t)} = 40 °C

Surrounding temperature, T_{s} = 20 °C

Using the equation of newton’s law of cooling,

– 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) we get,

(20/4) × (t/20) = 60/20

t/4 = 3

t = 12 minutes

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

**Problem 3:** A hot cup of soup is placed on the desk. The soup cools down from 100 °C to 80 °C in 6 minutes. If the temperature of the surroundings is 24 °C, then calculate the time taken by the soup to cool down from 70 °C to 50 °C.

Solution:**Case 1:** When the soup cools down from 100 °C to 80 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 20/6

Average temperature, T_{(t)} = 90 °C

Surrounding temperature, T_{s} = 24 °C

Using the equation of newton’s law of cooling,

– 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 °C to 50 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 20/t

Average temperature, T_{(t)} = 60 °C

Surrounding temperature, T_{s} = 24 °C

Using the equation of newton’s law of cooling,

– 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) we get,

(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 °C to 50 °C.

**Problem 4:** One large container contains some oil which has a temperature of 150 °C. The oil cools down to 120 °C in 8 minutes. What time will the oil take to cool down from 100 °C to 70 °C? The temperature of the surroundings is 23 °C.

Solution:**Case 1:** When the oil cools down from 150 °C to 120 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 30/8

Average temperature, T_{(t)} = 135 °C

Surrounding temperature, T_{s} = 23 °C

Using the equation of newton’s law of cooling,

– 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 °C to 70 °C

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

From the above statement… we get:

Rate of cooling, dQ/dt = 30/t

Average temperature, T_{(t)} = 85 °C

Surrounding temperature, T_{s} = 23 °C

Using the equation of newton’s law of cooling,

– 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) we get,

(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 °C to 70 °C.

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