# Problems on Newton’s Law of Cooling

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### Newton’s Law of Cooling:

• The rate of loss of heat by a body is directly proportional to its excess temperature over that of the surroundings provided that this excess is small.
• Let θ and θo, be the temperature of a body and its surroundings respectively. Let dQ / dt be the rate of loss of heat, So from Newton’s Law of Cooling, where k is a constant. Some times constant is denoted by C

#### Example – 1:

• A metal sphere, when suspended in a constant temperature enclosure, cools from 80 °C to 70 °C in 5 minutes and to 620C in the next five minutes. Calculate the temperature of the enclosure.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 80 °C to 70 °C : Initial temperature = θ1 = 80 °C, Final temperature = θ2 = 70 °C, Time taken t = 5 min

By Newton’s law of cooling • Consider a cooling from 70 °C to 62°C : Initial temperature = θ1 = 70 °C, Final temperature = θ2 = 62 °C, Time taken t = 5 min Dividing equation (1) by (2) 132 – 2 θo = 120 – 1.6θo

12 = 0.4 θo

θo = 12/0.4 = 30 °C

Ans: Surrounding temperature is 30 °C.

#### Example – 2:

• A metal sphere cools at the rate of 3 °C per minute when its temperature is 50 °C. Find its rate of cooling at 40 °C if the temperature of surroundings is 25 °C.
• Solution:
• Consider the cooling when temperature was 50 °C: Rate of cooling (dθ/dt)1= 3 °C per minute, temperature of body = θ1 = 50°C, temperature of surroundings = θo = 25 °C

By Newton’s law of cooling • Consider the cooling when temperature was 40 °C: temperature of body = θ1 = 40 °C, temperature of surroundings = θo = 25 °C, Rate of cooling (dθ/dt)2= ? Ans: The rate of cooling at 40 °C is 1.8 °C per minute,

#### Example – 3:

• A body cools at the rate of 0.5 °C/s when it is 500C above the surroundings. What is the rate of cooling when it is at 30 °C above the same surroundings?
• Solution:
• Consider the cooling when the temperature is 50 °C above the surroundings: Rate of cooling (dθ/dt)1= 0.5 °C per second, the temperature of the body above surroundings = (θ1 – θo)= 50 °C,

By Newton’s law of cooling • Consider the cooling when the temperature is 30 °C above the surroundings: temperature of the body above surroundings = (θ1 – θo)= 30 °C, Rate of cooling (dθ/dt)2= ? Ans: The rate of cooling at 30 °C above the  surroundings is1.8 °C per minute,

#### Example – 4:

• A metal sphere cools at the rate of 0.6 °C per minute when its temperature is 30 °C above surroundings. At what rate will it cool when its temperature is 20 °C above surroundings, other conditions remaining constant?
• Solution:
• Consider the cooling when the temperature is 30 °C above the surroundings: Rate of cooling (dθ/dt)1= 0.6 °C per minute, the temperature of the body above surroundings = (θ1 – θo)= 30 °C,

By Newton’s law of cooling • Consider the cooling when the temperature is 20 °C above the surroundings: temperature of the body above surroundings = (θ1 – θo)= 20 °C, Rate of cooling (dθ/dt)2= ? Ans: The rate of cooling at 20 °C above the  surroundings is 0.4 °C per minute,

#### Example – 5:

• A body at 50 °C cools in surroundings at 30 °C. At what temperature will its rate of cooling be half that at the beginning?
• Solution:

Temperature of surroundings = θo = 30 °C

• Consider the cooling when the temperature was θ1= 50 °C

By Newton’s law of cooling • Consider the cooling when the temperature was θ1 °C: Rate of cooling (dθ/dt)2= ½ (dθ/dt)1 Ans: at 40 °C the rate of cooling be half that at the beginning

#### Example – 6:

• A body cools from 75 °C to 55 °C in ten minutes when the surrounding temperature is 31°C. At what average temperature will its rate of cooling be ¼ th that at the start?
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 75 °C to 55 °C : Initial temperature = θ1 = 75 °C, Final temperature = θ2 = 55 °C, Time taken t = 10 min • Consider the cooling when the temperature was θ1 °C: Rate of cooling (dθ/dt)2= 1/4 (dθ/dt)1 Ans:

Ans: At temperature 39.5 °C the rate of cooling be ¼ th that at the start

#### Example – 7:

• A body cools from 60 °C to 50 °C in 5 minutes. How much time will it take to cool from 50 °C to 44 °C if the surrounding temperature is 32 °C?
• Solution:

Let θo = 44 °C  be the temperature of surroundings.

• Consider a cooling from 60 °C to 50 °C : Initial temperature = θ1 = 60 °C, Final temperature = θ2 = 50 °C, Time taken t = 5 min

By Newton’s law of cooling • Consider a cooling from 50 °C to 44 °C : Initial temperature = θ1 = 50 °C, Final temperature = θ2 = 44 °C Ans: Time taken to cool from 50 °C to 44 °C is 4.6 min

#### Example – 8:

• A body cools from 72 °C to 60 °C in 10 minutes. How much time will it take to cool from 60 °C to 52 °C if the temperature of surroundings is 36 °C?
• Solution:

Surrounding temperature = θo = 36 °C

• Consider a cooling from 72 °C to 60 °C: Initial temperature = θ1 = 72 °C, Final temperature = θ2 = 60 °C, Time taken t = 10 min

By Newton’s law of cooling • Consider a cooling from 60 °C to 52 °C : Initial temperature = θ1 = 60 °C, Final temperature = θ2 = 52 °C Ans: The time taken to cool from 60 °C to 52 °C is 10 min

#### Example – 9:

• A body cools from 750C to 70 °C in 2 minutes. What will additional time it take to cool to 60 °C, if the room temperature is 30 °C?
• Solution:

Surrounding temperature = θo = 30 °C

• Consider a cooling from 75 °C to 70 °C : Initial temperature = θ1 = 75 °C, Final temperature = θ2 = 70 °C, Time taken t = 2 min • Consider a cooling from 70 °C to 60 °C : Initial temperature = θ1 = 70 °C, Final temperature = θ2 = 60 °C Ans: The time taken to cool from 70 °C to 60 °C is 34/7 min

#### Example – 10:

• A heated metal ball is placed in cooler surroundings. Its rate of cooling is 2 °C per minute when its temperature is 60 °C and 1.2 °C per minute when its temperature is 52 °C. Find the temperature of the surroundings and the rate of cooling when the temperature of the ball is 48 °C.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling at 60 °C: temperature = θ1 = 60 °C, Rate of cooling = 2 °C per minute • Consider a cooling at 52 °C: temperature = θ2 = 52 °C, Rate of cooling = 1.2 °C per minute Dividing equation (1) by (2) ∴   52 – θo = 36 – 0.6 θo

∴   16 = 0.4 θo

∴   θo = 40 °C

substituting in equation (1)

2 = C (60 – 40)

∴   C = 1/10 min-1

Consider a cooling at 48 °C: temperature = θ3 = 48 °C, Ans: Temperature of surrounding is 40 °C and rate of cooling at 48 °C is 0.8 °C per min

#### Example – 11:

• A copper ball cools from 62 °C to 50 °C in 10 minutes and to 42 °C in the next 10 minutes. Calculate the temperature at the end of next 10 minutes.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 62 °C to 50 °C: Initial temperature = θ1 = 62 °C, Final temperature = θ2 = 50 °C, Time taken t = 10 min

By Newton’s law of cooling • Consider a cooling from 50 °C to 42 °C: Initial temperature = θ1 = 50 °C, Final temperature = θ2 = 42 °C, Time taken t = 10 min Dividing equation (2) by (1) ∴    138 – 3θo = 112 – 2θo

∴   θo  =26 oC

Surrounding temperature is 26 oC

substituting in equation (1)

1.2 = C( 56 -26)

∴    C = 1.2/30 = 1/25 min-1

• Consider further cooling from 42 oC to θ2 oC: Initial temperature = θ1 = 50 oC, Final temperature = θ2, Time taken t = 10 min #### Example – 12:

• A body cools from 60 oC to 52 oC in 5 minutes and from 52 oC to 44 oC in next 7.5 minutes. Determine its temperature in the next 10 minutes.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 60 oC to 52 oC : Initial temperature = θ1 = 60 oC, Final temperature = θ2 = 52 oC, Time taken t = 5 min

By Newton’s law of cooling • Consider a cooling from 52 oC to 44 oC : Initial temperature = θ1 = 52 oC, Final temperature = θ2 = 42 oC, Time taken t = 7.5 min Dividing equation (2) by (1) ∴    72 – 1.5 θo = 56 – θo

∴    16 = 0.5 θo

∴    θo = 16/0.5 = 32 oC

Surrounding temperature is 32 oC

substituting in equation (1)

1.6 = C( 56 -32)

C = 1.6/24 = 1/15 min-1.

• Consider further cooling from 44 oC to θ2 oC : Initial temperature = θ1 = 44 oC, Final temperature = θ2, Time taken t = 10 min Ans: The temperature after 10 minutes is 38 oC

#### Examples – 13:

• A body cools from 60 oC to 52 oC in 10 minutes and to 46 oC in next 10 minutes. Find the temperature of surroundings.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 60 oC to 52 oC : Initial temperature = θ1 = 60 oC, Final temperature = θ2 = 52 oC, Time taken t = 10 min

By Newton’s law of cooling • Consider a cooling from 52 oC to 46 oC : Initial temperature = θ1 = 52 oC, Final temperature = θ2 = 46 oC, Time taken t = 10 min Dividing equation (2) by (1) ∴    196 – 4θo = 168 – 3θo

∴     θ= 28

Ans: Surrounding temperature is 28 oC

#### Example – 14:

• The rate of cooling of a body is 2 oC/min at temperature 60 oC and 1 oC/min at 45 oC. What will be the temperature of the surroundings?
• Solution:

Let θo be the temperature of surroundings.

• Consider the cooling when temperature θ1 = 60 oC: Rate of cooling (dθ/dt)1= 2 oC per minute,

By Newton’s law of cooling • Consider the cooling when temperature θ2 = 30 oC: Rate of cooling (dθ/dt)1= 1 oC per minute, Dividing equation (1) by (2) ∴  90 – 2θo= 60 – θo

∴  θ= 30 oC

Ans: Surrounding temperature is 30oC

#### Example – 15:

• A metal sphere cools from 60 oC to 50 oC in 5 minutes. How much more time will it take to cool from 50 oC to 40 oC if the temperature of the surroundings is 30 oC.
• Solution:

Let θo be the temperature of surroundings.

• Consider a cooling from 60 oC to 50 oC : Initial temperature = θ1 = 60 oC, Final temperature = θ2 = 50 oC, Time taken t = 5 min

By Newton’s law of cooling • Consider a cooling from 50 oC to 40 oC : Initial temperature = θ1 = 50 oC, Final temperature = θ2 = 40 oC, Time taken t = ? Ans: Time taken to cool from 50 oC to 40 oC is 8.33 min

#### Example – 16:

• A copper sphere is heated and then allowed to cool while suspended in an enclosure whose walls are maintained at a constant temperature. When the temperature of the sphere is 86 oC, it is cooling at the rate of 3 oC/min; at 75 oC, it is cooling at the rate of 2.5 oC/min. What is the temperature of the sphere when it is cooling at the rate of 1 oC/min? Assume Newton’s law of cooling.
• Solution:

Let θo be the temperature of surroundings.

• Consider the cooling when temperature θ1 = 86 oC: Rate of cooling (dθ/dt)1= 3 oC per minute,

By Newton’s law of cooling • Consider the cooling when temperature θ1 = 75 oC: Rate of cooling (dθ/dt)1= 2.5 oC per minute, Dividing equation (2) by (1) ∴  450 – 6θo = 430 – 5 θo

∴  θ= 20 oC

Substituting in equation (1)

3 = C(86 – 20)

∴   C = 3/66 = 1/22 min-1

• Consider the cooling when temperature = θ3: Rate of cooling (dq/dt)1=  1 oC per minute, ans: At 42 oC it is cooling at the rate of 1 oC/min

 Science > You are Here

1. Nitish sachan

Awesom and easy question

2. Sujit More

i want solution of this example:-
Water at temperature 100 degree celsius cools in 10 minutes to 88 degree celsius in a room of temperature 25 degree celsius. Find the temperature of water after 20 minutes.

• Hemant More

Part – I
Consider a cooling from 100 oC to 88 oC : Initial temperature = θ1 = 100 oC, Final temperature = θ2 = 88 oC, Time taken t = 10min, surrounding temperature θo = 25 oC,
θ = (θ1 + θ2)/2 = (100 + 88)/2 = 188/2 = 94 oC
dθ/dt = (θ1 – θ2)/dt = (100 – 88)/10 = 12/10 = 1.2 oC/min
By Newton's law of cooling
dθ/dt = C(θ – θo)
1.2 = C(94 – 25)
1.2 = 69 C
C = 1.2/69 per min
Part – II
Consider cooling for next 20 minutes
Initial temperature = θ1 = 88 oC, Final temperature = θ2 = θ, Time taken t = 20min, surrounding temperature θo = 25 oC,
θ = (θ1 + θ2)/2 = (100 + 88)/2 = 188/2 = 94 oC
dθ/dt = (θ1 – θ2)/dt = (100 – 88)/2 = 12/2 = 2 oC/min
By Newton's law of cooling
dθ/dt = C(θ – θo)
(θ1 – θ2)/dt = C((θ1 + θ2)/2) – θo)
(88 – θ)/20 = (1.2/69)((88 + θ)/2) – 25)
4.4 – 0.05 θ = (1.2/69)(44 + 0.5 θ – 25)
69(4.4 – 0.05 θ) = 1.2(19 + 0.5 θ)
303.6 – 3.45 θ = 22.8 + 0.6 θ
303.6 – 22.8 = 3.45 θ + 0.6 θ
280.8 = 4.05 θ
θ = 280.8/4.05 = 69.33 oC
Note: If Time is from start then in part – II θ1 = 100 oC

3. keerthana

Good examples

4. LAV PRASAD YADAV

wELL DONE sIR,
A MOST INFORMATIVE AND CONCEPTUAL WORK FROM A BEST TEACHERS ,SO FAR I FEEL.

5. Shaikh Fardin

I have one question
A body cools from 70°c to 60°c in 5 minutes and to 45°c in next 10 minutes
calculate the temperature of surroundings?

• Hemant More

It is similar to solved example – 01