NCERT Solutions
Class 11 Physics
Thermal Properties of Matter
Q. 10.20
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.
According to Newton’s law of cooling, we have:
Where
(-dT/dt) = K (T – To)
(dt)/ (K (T – To) = -K dt
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:
30∫80 (dt)/ (K (T – To)) =0∫300 K dt
[Loge (T – To (50)) 80 = -K[t] 0300
(2.3026)/ (K) Log10 ((80-20)/ (50-20)) =-300
(-2.3026)/ (300) log10 2 =K ... (ii)
The temperature of the body falls from 60°C to 30°C in time = t’
Hence we get:
(2.3026)/ (K) Log10 ((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) log10 2
t' = (300 x 2) = 600s = 10min
Therefore, the time taken to cool the body from 60°C to 30°C is 10 minutes.