An object kept in a large room having air temperature of 25°C takes 12 minutes to cool from 80° to 70°C. The time taken to cool for the same object from 70°C to 60°C would be nearly :
- 10 min
- 12 min
- 20 min
- 15 min
The Correct Option is D
Solution and Explanation
\(\frac{80 - 70}{12} = K( \frac{80 + 70}{2} - 25) \,...(1)\)
\(\frac{70 - 60}{t} = K(\frac{70 + 60}{2} - 25) \,...(2)\)
on solving : \(t = 15\) min
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Concepts Used:
Newton’s Law of Cooling
Newton’s law of cooling states that the rate of heat loss from a body is directly proportional to the difference in temperature between the body and its surroundings.
Derivation of Newton’s Law of Cooling
Let a body of mass m, with specific heat capacity s, is at temperature T2 and T1 is the temperature of the surroundings.
If the temperature falls by a small amount dT2 in time dt, then the amount of heat lost is,
dQ = ms dT2
The rate of loss of heat is given by,
dQ/dt = ms (dT2/dt) ……..(2)
Compare the equations (1) and (2) as,
– ms (dT2/dt) = k (T2 – T1)
Rearrange the above equation as:
dT2/(T2–T1) = – (k / ms) dt
dT2 /(T2 – T1) = – Kdt
where K = k/m s
Integrating the above expression as,
loge (T2 – T1) = – K t + c
or
T2 = T1 + C’ e–Kt
where C’ = ec