the temperature of a body falls from 30C to 20C in 5 minutes. The air temperature is 13C. Find the temperature after a further 5 minutes.
T(s) = 13C is the air temperature,
the initial temperature T(o) = 30C.
Newton’s law of cooling is differential equation
dT/dt = - k(T-T(s))
Its solution is
T =T(s) + (T(o) - T(s)) •e^-kt.
After the first 5 min
20 = T(s) + (30 – T(s)) •e^-5•k. (1)
After a further 5 min, therefore, after 10 minutes
T = T(s) + (30 – T(s)) •e^-10•k. (2)
From the equation (1)
e^-5•k = (20 -T(s))/(30-T(s))
From the equation (2)
e^-10•k = (e^-5•k)^2 = [(20 -T(s))/(30-T(s))]^2.
Then the equation (2) is
T = T(s) + (30 – T(s)) •e^-10•k =
= T(s) + (30 – T(s)) • [(20 -T(s))/(30-T(s))]^2=
= T(s) + (20 –T(s))^2/(30-T(s) =
Adam
To find the temperature after a further 5 minutes, we need to understand the rate of change of temperature.
The temperature of the body is falling at a constant rate of (30C - 20C) / 5 minutes = 2C per minute.
Since the air temperature is constant at 13C, we can assume that the body will continue to cool at the same rate of 2C per minute.
After a further 5 minutes, the body will have cooled for a total of 10 minutes (initial 5 minutes + further 5 minutes).
In 10 minutes, the body will have cooled by 10 minutes x 2C per minute = 20C.
Therefore, the temperature after a further 5 minutes will be:
20C - 20C = 0C.
So, the temperature of the body after a further 5 minutes will be 0C.