# First order ordinary differential equation

1. Jun 17, 2011

### KillerZ

1. The problem statement, all variables and given/known data

I haven't done ODEs in a few years and I am trying to do this equation:

$m_{Hg}C_{p,Hg}\frac{dT_{Hg}}{dt} = Q$

$Q = hA(T_{air} - T_{Hg})$

$T_{Hg}(t = 0) = 20$

I need to find $T_{Hg}(t=590)$

2. Relevant equations

3. The attempt at a solution

$h, A, m_{Hg}, C_{p,Hg}, T_{air}$ are all constants

$\frac{dT_{Hg}}{dt} = \frac{hA(T_{air} - T_{Hg})}{m_{Hg}C_{p,Hg}}$

$\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = dt$

$\int\frac{m_{Hg}C_{p,Hg}}{hA(T_{air} - T_{Hg})} dT_{Hg} = \int dt$

$\frac{-m_{Hg}C_{p,Hg}}{hA}\int\frac{1}{(-T_{air} + T_{Hg})} dT_{Hg} = \int dt$

$\frac{-m_{Hg}C_{p,Hg}}{hA} ln|T_{Hg} - T_{air}| = t + c$

$ln|T_{Hg} - T_{air}| = \frac{t + c}{\frac{-m_{Hg}C_{p,Hg}}{hA}}$

$T_{Hg} - T_{air} = e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}}$

$e^{c/\frac{-m_{Hg}C_{p,Hg}}{hA}} = c$

$T_{Hg} - T_{air} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}}$

$T_{Hg} = ce^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}$

$T_{Hg}(t = 0) = 20$

$T_{air} = -7$

$h = 3.9$

$A = 0.00176$

$C_{p,HG} = 139.0908$

$m_{Hg} = 0.05$

$20 = ce^{0/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)$

$c = 27$

$T_{Hg} = 27e^{t/\frac{-m_{Hg}C_{p,Hg}}{hA}} + T_{air}$

$T_{Hg}(t=590)$

$T_{Hg}(t=590) = 27e^{590/\frac{-(0.05)(139.0908)}{(3.9)(0.00176)}} + (-7)$

$T_{Hg}(t=590) = 8.0822$

2. Jun 17, 2011

### LCKurtz

Man, you must like typing subscripts. Why not call your dependent variable T and write the equation in the form

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

in the first place and put the constants in for k at the end? It would be much easier to read, not to mention typing it all in. Anyway, it looks like you solved it OK, with the caveat that I didn't check all the details.

3. Jun 17, 2011

### hunt_mat

I think you made an error in using your initial condition (using the notation of LCKurtz) you had:
$$-\ln |T_{a}-T|=kt+c$$
Now if $T(0)=T_{0}$, then I would use that at this point here to tell you that $-\ln |T_{a}-T_{0}|=c$ and your equation becomes:
$$-\ln |T_{a}-T|=kt-\ln |T_{a}-T_{0}|$$
From here I think it is just algebra to obtain the correct answer.

4. Jun 17, 2011

### KillerZ

OK doing it with the substitutions:

$k = \frac{hA}{m_{Hg}C_{p,Hg}}$

and

$T = T_{Hg}$

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

$\frac{1}{k(T_a - T)}dT = dt$

$\frac{-1}{k}\int\frac{1}{(-T_a + T)}dT = \int dt$

$\frac{-1}{k}ln|-T_a + T| = t + c$

$ln|-T_a + T| = -kt + c$

$-T_a + T = e^{-kt + c}$

$T = ce^{-kt} + T_a$

Then initial conditions T(t=0) = 20

$20 = ce^{-k(0)} + (-7)$

$27 = c$

$T = 27e^{-kt} + T_a$

Last edited: Jun 17, 2011
5. Jun 17, 2011

### hunt_mat

This line is incorrect:
$$T=ce^{-kt}+T_{a}$$