# 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 - Physics

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 1

According to Newton’s law of cooling, we have:

- (dT)/(dt) = K(T - T_0)

(dT)/(K(T-T_0)) = -kdt ...(i)

Where,

Temperature of the body = T

Temperature of the surroundings = T0 = 20°C

is a constant

Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s

Integrating equation (i), we get:

int_50^80 (dt)/(K(T - T_0)) = -int_0^300Kdt

[log_e(T-T_0)]_50^80 = -K[t]_0^300

2.3026/K log_10 (80-20)/(50-20) = -300

2.3026/K =log_10 2 = -300   ....(ii)

The temperature of the body falls from 60°C to 30°C in time = t

Hence, we get:

2.3026/K log_10  (60-20)/(30-20) = -t

-2.3026/t log_10 4 = K ...(iii)

Equating equations (ii) and (iii), we get:

-2.3026/t log_10 4 = (-2.3026)/300 log_10 2

:.t  = 300 x 2 = 600 s = 10 min

Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.

#### Solution 2

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

Here Average of 80 ^@C and 50 ^@C = 65 ^@C

Temperature of surroundings = 20^@C

:. Difference = 65 - 20 = 45 ^@C

Under these condition. the body cools 30^@C in time 5 minutes

:.  "Change in temp"/"Time" = K triangleT or 30/5 = K xx 45^@  .. (1)

The average of  60^@C and 30^@ is 45^@C which is 25^@C(45 - 20) above the room temperature anf the bodycppls by 30^@C(60 - 30) in time t (say)

:. 30/t =  K xx 25   ...(ii)

Where K is same for this situation as for the original.

Dividing equation i by ii we get

="30/5"/"30/t" = (Kxx45)/(Kxx25)

or t/5 = 9/5

=> t = 9 min`

Concept: Newton’s Law of Cooling
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#### APPEARS IN

NCERT Class 11 Physics
Chapter 11 Thermal Properties of Matter
Exercises | Q 22 | Page 297