**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.,

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

**dT/dt = k(T****t ****â€“ T****s****)**

Where

**T****t** = temperature at time t and

** T****s** = temperature of the surrounding,

** Â k** = constant.

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

**T(t) = T****s**** + (T****0**** â€“ T****s****) e****-Kt**

Where,

t = time,

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

**T****s** = surrounding temperature,

**T****o** = 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.