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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 θ
_{o}*,*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 θ_{o} be the temperature of surroundings.

- Consider a cooling from 80 °C to 70 °C : Initial temperature = θ
_{1}= 80 °C, Final temperature = θ_{2}= 70 °C, Time taken t = 5 min

By Newton’s law of cooling

- Consider a cooling from 70 °C to 62°C : Initial temperature = θ
_{1}= 70 °C, Final temperature = θ_{2}= 62 °C, Time taken t = 5 min

Dividing equation (1) by (2)

132 – 2 θ_{o }= 120 – 1.6θ_{o}

12 = 0.4 θ_{o}

θ_{o }= 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 = θ_{1}= 50°C, temperature of surroundings = θ_{o}= 25 °C

By Newton’s law of cooling

- Consider the cooling when temperature was 40 °C: temperature of body = θ
_{1}= 40 °C, temperature of surroundings = θ_{o}= 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 = (θ_{1}– θ_{o})= 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 = (θ
_{1}– θ_{o})= 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 = (θ_{1}– θ_{o})= 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 = (θ
_{1}– θ_{o})= 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 = θ_{o} = 30 °C

- Consider the cooling when the temperature was θ
_{1= }50 °C

By Newton’s law of cooling

- Consider the cooling when the temperature was θ
_{1}°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 θ_{o} be the temperature of surroundings.

- Consider a cooling from 75 °C to 55 °C : Initial temperature = θ
_{1}= 75 °C, Final temperature = θ_{2}= 55 °C, Time taken t = 10 min

- Consider the cooling when the temperature was θ
_{1}°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 θ_{o} = 44 °C be the temperature of surroundings.

- Consider a cooling from 60 °C to 50 °C : Initial temperature = θ
_{1}= 60 °C, Final temperature = θ_{2}= 50 °C, Time taken t = 5 min

By Newton’s law of cooling

- Consider a cooling from 50 °C to 44 °C : Initial temperature = θ
_{1}= 50 °C, Final temperature = θ_{2}= 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 = θ_{o} = 36 °C

- Consider a cooling from 72 °C to 60 °C: Initial temperature = θ
_{1}= 72 °C, Final temperature = θ_{2}= 60 °C, Time taken t = 10 min

By Newton’s law of cooling

- Consider a cooling from 60 °C to 52 °C : Initial temperature = θ
_{1}= 60 °C, Final temperature = θ_{2}= 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 = θ_{o} = 30 °C

- Consider a cooling from 75 °C to 70 °C : Initial temperature = θ
_{1}= 75 °C, Final temperature = θ_{2}= 70 °C, Time taken t = 2 min

- Consider a cooling from 70 °C to 60 °C : Initial temperature = θ
_{1}= 70 °C, Final temperature = θ_{2}= 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 θ_{o} be the temperature of surroundings.

- Consider a cooling at 60 °C: temperature = θ
_{1}= 60 °C, Rate of cooling = 2 °C per minute

- Consider a cooling at 52 °C: temperature = θ
_{2}= 52 °C, Rate of cooling = 1.2 °C per minute

Dividing equation (1) by (2)

∴ 52 – θ_{o} = 36 – 0.6 θ_{o}

∴ 16 = 0.4 θ_{o}

∴ θ_{o} = 40 °C

substituting in equation (1)

2 = C (60 – 40)

∴ C = 1/10 min^{-1}

Consider a cooling at 48 °C: temperature = θ_{3} = 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 θ_{o} be the temperature of surroundings.

- Consider a cooling from 62 °C to 50 °C: Initial temperature = θ
_{1}= 62 °C, Final temperature = θ_{2}= 50 °C, Time taken t = 10 min

By Newton’s law of cooling

- Consider a cooling from 50 °C to 42 °C: Initial temperature = θ
_{1}= 50 °C, Final temperature = θ_{2}= 42 °C, Time taken t = 10 min

Dividing equation (2) by (1)

∴ 138 – 3θ_{o} = 112 – 2θ_{o}

∴ θ_{o} =26 ^{o}C

Surrounding temperature is 26 ^{o}C

substituting in equation (1)

1.2 = C( 56 -26)

∴ C = 1.2/30 = 1/25 min^{-1}

- Consider further cooling from 42
^{o}C to θ_{2}^{o}C: Initial temperature = θ_{1}= 50^{o}C, Final temperature = θ_{2}, Time taken t = 10 min

#### Example – 12:

- A body cools from 60
^{o}C to 52^{o}C in 5 minutes and from 52^{o}C to 44^{o}C in next 7.5 minutes. Determine its temperature in the next 10 minutes. **Solution:**

Let θ_{o} be the temperature of surroundings.

- Consider a cooling from 60
^{o}C to 52^{o}C : Initial temperature = θ_{1}= 60^{o}C, Final temperature = θ_{2}= 52^{o}C, Time taken t = 5 min

By Newton’s law of cooling

- Consider a cooling from 52
^{o}C to 44^{o}C : Initial temperature = θ_{1}= 52^{o}C, Final temperature = θ_{2}= 42^{o}C, Time taken t = 7.5 min

Dividing equation (2) by (1)

∴ 72 – 1.5 θ_{o} = 56 – θ_{o}

∴ 16 = 0.5 θ_{o}

∴ θ_{o} = 16/0.5 = 32 ^{o}C

Surrounding temperature is 32 ^{o}C

substituting in equation (1)

1.6 = C( 56 -32)

C = 1.6/24 = 1/15 min^{-1}.

- Consider further cooling from 44
^{o}C to θ_{2}^{o}C : Initial temperature = θ_{1}= 44^{o}C, Final temperature = θ_{2}, Time taken t = 10 min

Ans: The temperature after 10 minutes is 38 ^{o}C

#### Examples – 13:

- A body cools from 60
^{o}C to 52^{o}C in 10 minutes and to 46^{o}C in next 10 minutes. Find the temperature of surroundings. **Solution:**

Let θ_{o} be the temperature of surroundings.

- Consider a cooling from 60
^{o}C to 52^{o}C : Initial temperature = θ_{1}= 60^{o}C, Final temperature = θ_{2}= 52^{o}C, Time taken t = 10 min

By Newton’s law of cooling

- Consider a cooling from 52
^{o}C to 46^{o}C : Initial temperature = θ_{1}= 52^{o}C, Final temperature = θ_{2}= 46^{o}C, Time taken t = 10 min

Dividing equation (2) by (1)

∴ 196 – 4θ_{o} = 168 – 3θ_{o}

∴ θ_{o }= 28

**Ans:** Surrounding temperature is 28 ^{o}C

#### Example – 14:

- The rate of cooling of a body is 2
^{o}C/min at temperature 60^{o}C and 1^{o}C/min at 45^{o}C. What will be the temperature of the surroundings? **Solution:**

Let θ_{o} be the temperature of surroundings.

- Consider the cooling when temperature θ
_{1}= 60^{o}C: Rate of cooling (dθ/dt)_{1}= 2^{o}C per minute,

By Newton’s law of cooling

- Consider the cooling when temperature θ
_{2}= 30^{o}C: Rate of cooling (dθ/dt)_{1}= 1^{o}C per minute,

Dividing equation (1) by (2)

∴ 90 – 2θ_{o}= 60 – θ_{o}

∴ θ_{o }= 30 ^{o}C

**Ans:** Surrounding temperature is 30^{o}C

#### Example – 15:

- A metal sphere cools from 60
^{o}C to 50^{o}C in 5 minutes. How much more time will it take to cool from 50^{o}C to 40^{o}C if the temperature of the surroundings is 30^{o}C. **Solution:**

Let θ_{o} be the temperature of surroundings.

- Consider a cooling from 60
^{o}C to 50^{o}C : Initial temperature = θ_{1}= 60^{o}C, Final temperature = θ_{2}= 50^{o}C, Time taken t = 5 min

By Newton’s law of cooling

- Consider a cooling from 50
^{o}C to 40^{o}C : Initial temperature = θ_{1}= 50^{o}C, Final temperature = θ_{2}= 40^{o}C, Time taken t = ?

**Ans:** Time taken to cool from 50 ^{o}C to 40 ^{o}C 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
^{o}C, it is cooling at the rate of 3^{o}C/min; at 75^{o}C, it is cooling at the rate of 2.5^{o}C/min. What is the temperature of the sphere when it is cooling at the rate of 1^{o}C/min? Assume Newton’s law of cooling. **Solution:**

Let θ_{o} be the temperature of surroundings.

- Consider the cooling when temperature θ
_{1}= 86^{o}C: Rate of cooling (dθ/dt)_{1}= 3^{o}C per minute,

By Newton’s law of cooling

- Consider the cooling when temperature θ
_{1}= 75^{o}C: Rate of cooling (dθ/dt)_{1}= 2.5^{o}C per minute,

Dividing equation (2) by (1)

∴ 450 – 6θ_{o} = 430 – 5 θ_{o}

∴ θ_{o }= 20 ^{o}C

Substituting in equation (1)

3 = C(86 – 20)

∴ C = 3/66 = 1/22 min^{-1}

- Consider the cooling when temperature = θ
_{3}: Rate of cooling (dq/dt)_{1}= 1^{o}C per minute,

ans: At 42 ^{o}C it is cooling at the rate of 1 ^{o}C/min

Science > Physics > Radiation > You are Here |

Physics |
Chemistry |
Biology |
Mathematics |

Awesom and easy question

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.

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

Good examples