Continuous Compounding Formula Newtons Law of Cooling

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.

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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.

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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.

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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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Related:

• Newton's Second Law Equation
• Newton's Law of Universal Gravitation Formula
• Coulomb's Law Equation
• Snell's Law Equation
• Ohm's Law Equation
• Hooke's Law Equation

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