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Question
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
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:
`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`
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