11.20 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.
11.20 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.
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1 Answer
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11.20 According to Newton’s law of cooling, we have
= K(T - ) orT o = -Kdt……(i)Where, temperature of the body = Td T K ( T - T o ) Temperature of the surroundings =
= 20 °CT o K is a constant
The temperature of the body falls from 80
to 50? ? i n t i m e t = 5 min ? = 300 s Integrating equation (i), we get
= -50 80 d T K ( T - T o ) ∫ 0 300 K d t∫ = -K? log e ? ( T - T o ) 30 80 t 0 300 = -300? 2.3026 K log 10 ? 80 - 20 50 - 20 = K ….(ii)? 2.3026 300 log 10 ? 2 The temperature of the body falls from 60
to 30? ? i n t i m e = t ' Hence, we get:
= -t'2.3026 K log 10 ? 60 - 20 30 - 20 = K.(iii)? - 2.3026 t ' log 10 ? 4 Equating equations (ii) and (iii), we get,
=2.3026 300 log 10 ? 2 - 2.3026 t ' log 10 ? 4 = 10 min? t ' = 300 × 2 = 600 s Hence, the time taken to cool the body from 60
to 30? is 10 minutes.?
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