**Newton’s Law of Cooling Formula**

Newton’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 object’s temperature and its surroundings, provided the difference is small, i.e.,

The Newton’s law of cooling is given by,

**dT/dt = k(T****t ****– T****s****)**

Where

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

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

** k** = constant.

The **Newton’s Law of Cooling Formula** is expressed by

**T(t) = T****s**** + (T****0**** – 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)−Ts / To−Ts = e-kt

-kt ln = ln T(t)−Ts / To−Ts

-kt = ln 50−25 / 70−25

= 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)−Ts / To−Ts

– (0.092) t = ln 45−25 / 70−25

– 0.092 t = -0.597

t = −0.597 / −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

= 25 + (80 – 25)e-(0.56)

= 25 + 55 × 0.57

= 45.6 oC

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