Question 11.22:
(-dT/T) = K(T - T0)
dT / K(T - T0) = -Kdt ....(i)
Where,
Temperature of the body = T
Temperature of the surroundings = T0 = 20°C
K 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:
The
temperature of the body falls from 60°C
to 30°C in time = t’
Hence, we get:
∴ 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.
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:
According
to Newton’s law of cooling, we have:(-dT/T) = K(T - T0)
dT / K(T - T0) = -Kdt ....(i)
Where,
Temperature of the body = T
Temperature of the surroundings = T0 = 20°C
K 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:

Hence, we get:

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