A cup of coffee cools from 90°C to 80°C in t minutes, when the room temperature is 20°C. The time taken by a similar cup of coffee to cool from 80°C to 60°C at a room temperature same at 20°C is
\(\frac{5}{13}t\)
\(\frac{13}{10}t\)
\(\frac{13}{5}t\)
\(\frac{10}{13}t\)
The Correct Option is C
Solution and Explanation
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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