# Newtons Law Of Cooling Formula

## Newton’s Law of Cooling Formula

Newtonâ€™s law of cooling describes the rate at which an exposed body changes temperature through radiation which is approximately proportional to the difference between the objectâ€™s temperature and its surroundings, provided the difference is small, i.e.,

$\frac{dT}{dt}\alpha&space;(T_{t}-&space;T_{s})$

The Newtonâ€™s law of cooling is given by,

dT/dt = k(Tt â€“ Ts)

Where

Tt = temperature at time t and

Ts = temperature of the surrounding,

Â k = constant.

The Newtonâ€™s Law of Cooling Formula is expressed by

T(t) = Ts + (T0 â€“ Ts) e-Kt

Where,

t = time,

T(t) = temperature of the given body at time t,

Ts = surrounding temperature,

To = initial temperature of the body,

k = constant.

Please note that greater the difference in temperature between the system and surrounding, more rapidly the body temperature changes.

Example 1

The oil is heated to 70oC. It cools to 50oC after 6 minutes. Calculate the time taken by the oil to cool from 50oC to 40oC given the Surrounding temperature Ts = 25oC.

Solution:

Given:

Temperature of oil after 10 min = 50oC,

Ts = 25oC,

To = 70oC,

t = 6 min

The Newton’s law of cooling formula is expressed as

T(t) = Ts + (To – Ts) e-kt

T(t)âˆ’Ts / Toâˆ’Ts = e-kt

-kt ln = ln T(t)âˆ’Ts / Toâˆ’Ts

-kt = ln 50âˆ’25 / 70âˆ’25

Â Â Â Â = ln 0.555

k = – (-0.555 / 6 )

= 0.092

If Tt = 45oC (average temperature as the temperature decreases from 50oC to 40oC)

Time taken is -kt ln e = ln T(t)âˆ’Ts / Toâˆ’Ts

– (0.092) t = ln 45âˆ’25 / 70âˆ’25

– 0.092 t = -0.597

t = âˆ’0.597 / âˆ’0.092

= 6.489 min.

Example 2: Anil heats the water to 80oC. He waits for 10 min. How much would be the temperature if k = 0.056 per min and surrounding temperature is 25oC.

Solution:

Given:

Ts = 25oC,

To = 80oC,

t = 10 min,

k = 0.056

The Newton’s law of cooling formula is expressed as

T(t) = Ts + (To – Ts)e-kt

Â Â Â Â Â = 25 + (80 – 25)e-(0.56)

Â Â Â Â Â = 25 + 55 Ã— 0.57 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â

Â  = 45.6 oC

Temperature cools down from 80oC to 45.6oC after 10 min.

#### Practise This Question

Animals in which the body cavity is not lined by mesoderm, and instead, the mesoderm is present as scattered pouches in between the ectoderm and endoderm are