Question

Thermometer is taken outside from a room, where the air temperature is 5F. After 1 min. the thermometer reads 55F and after 5mins it reads 30F. What is the initial temperature of the room?

Answer 1

\[\frac{ dP }{ dt } = k(T-T_m)\]

Answer 2

Solving the equation:
\[T(t)= T_m+ce^{kt}\]

Answer 3

Looks like we are looking for Tm

Answer 4

If i assume T(0) = 5F outside maybe \[5 = T_m+c\] or \[5-T_m = c\]

Answer 5

So at T(1) = 55
\[55 = T_m+(5-T_m)e^k\]
and at T(5) = 30
\[30 = T_m+(5-T_m)e^{5k}\]

Answer 6

System of equations?

Answer 7

presumably, you are supposed to use Newton's law of cooling, which says that the instantaneous rate of temperature change of a body is proportional to the difference in temperature between the body and the surroundings. In mathematical form, this is expressed as:

dT(t)/dt = k*(Tsurr - T(t))

where T(t) is the temperature of the body at time t, Tsurr is the temperature of the surroundings, and k is a positive constant of proportionality. I will assume that the temperature of the surroundings is constant, and that the thermometer was in thermal equilibrium with the room before being brought outside.

With assumption that Tsurr is a constant, the differential equation above is a separable equation:

dT/(Tsurr - T) = k dt

Integrating, we get:

-ln(Tsurr - T) = k*t - c

where c is the constant of integration.

T(t) = Tsurr + exp(c - k*t)

Remember that exp(a+b) = exp(a*exp(b), so we can write the solution as:

T(t) = 5F+ C*exp(-k*t)

where C = exp(c), and Tsurr = 5F

This is the general solution to the differential equation. It contains two unknown constants (C, and k) , but we are given two "facts" that will allow us to solve for these constants.

At t = 1 min, T(1 min) = 55F and at t = 5 min, T(5min) = 30 F, so using the data at 1 min, we have:

55F = 5F+ C*exp(-k*min)

50F*exp(k*min) = C

This defines C in terms of k.

Now using the second condition:

30F = 5F + C*exp(-5k*min)

Plug in the expression for C as a function of k:

25F = 50F*exp(k*min)*exp(-5k*min)

1/2 = exp(-4k*min)

ln(1/2) = -4k*min

(ln(2))/(4min) = k = 0.1733/min

So:

C = 50F*exp((ln(2))/4) = 59.460 F

The particular solution for this case is therefore:

T(t) = 5F+ (59.460F)*exp(-0.1733*t/min)

At t = 0, the temperature of the thermometer, and by implication, the temperature of the room, is:

T(0) = 5F + 59.460F = 64.46F