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.

Answer 1

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) =

Answer 2

Adam