A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Solution:

Given:

In this question, the initial temperature of the body is 80 degrees Celsius, and the final temperature of the body is 50 degrees Celsius.
(T1 T2)/2 is the average temperature.

Avg Temp= (80 + 50)/2 = 65 0C
Temperature of the surrounding, T0 =20 0C

Temperature difference, ΔT= 65 0C  – 20 0C = 45 0C
t=5 min
Newton’s law of cooling states that the rate of cooling is proportional to the difference in temperature between two points.

Change in temperature/ time = K ΔT

(T2 – T1)/t = KΔT
(80 – 50)/5 = K(45)

6 = K(45)

K = 6/45

K = 2/15

 

Average

In the second situation,
The body’s initial temperature, </span></span><span class=" jss3166 jss3228"><span id="editable-content-within-article~1~3~0" class="jss3168 jss3229">T_1, is 60 degrees Celsius.
T_2 is the last temperature of the body, and it is 30 degrees Celsius.
Let be the amount of time required for cooling.

temperature, (T1 + T2)/2 = (60 + 30)/2 = 45 0C

Temperature difference, ΔT = 45 0C – 20 0C = 25 0C

According to Newton’s law of cooling, the rate of cooling is proportional to the difference in temperature

Change in temperature/ time = K ΔT

(T2 – T1)/t = KΔT

(60 – 30)/t = (2/15) (25)

30/t = 3.33

t = 30/3.33 = 9 min

Thus, answer is 9 mins.