Question

An object is taken out of a 21 degree C room and placed outside where the temperature is 4 degree C. Twenty-five minutes later the temperature is 17 degree C. It cools according to Newton's Law. Determine the temperature of the object after one hour. Round to the nearest tenth of a degree.

Answer 1

Answer: The temperature of the object after one hour is 12.9°C to the nearest tenth of a degree

Step-by-step explanation:

Please see the attachments below

Answer 2

Answer:

12.9 degree C

Step-by-step explanation:

1 hour = 60 minutes

So the Newton's law of cooling state that the rate of changing in temperature is proportional to the different in temperature, in other words:

$\frac{dT}{dt} = -k(T - T_e)$

where t is the time in minute, k is the cooling constant and $T_e = 4^o C$ is the environmental temperature. This differential equation can be solved to have a form of

$T(t) = T_e + (T_0 - T_e)e^{-kt}$

where $T_0 = 21^o C$ is the initial temperature.

We know that at 25 minutes T(25) = 17

$4 + (21 - 4)e^{-25k} = 17$

$4 + 17e^{-25k} = 17$

$e^{-25k} = (17 - 4)/17 = 0.765$

$-25k = ln(0.765) = -0.268$

$k = -0.268 / -25 = 0.0107$

So after 60 minutes:

$T(60) = 4 + 17e^{-0.0107*60} = 4+17*0.525 = 12.9 ^o C$