Newton’s law of cooling to find the temperature of dead body

[songoku]

Homework Statement

Polices found dead body. At 1 pm when the pathologist arrived the temperature of the victim was 15◦C. It was a cold winter day and the weather office reported the temperature in the city at 6 am was −2◦C, but increasing at a rate of 1◦C per hour. Assume that the body was originally at a temperature of 37◦C. The pathologist determined from experiments that in a room at 5◦C the victim would cool from 15◦C to 12◦C in one hour.

Newton’s law of cooling states that a body at temperature T(t) will cool to an ambient temperature $T_{a}(t)$ at a constant rate, given by $\frac{dT}{dt}=-k(T-T_{a}(t))$

(a) Write down an expression for the atmospheric temperature $T_{a}(t)$, where t = 0 at 1 pm.
(b) Write down an ODE to model the experiment. Solve this equation and use this to determine the value of $k$.
(c) Write down an ODE to model the environment in which the murder took place (you will need to use part (a)). Solve this equation and determine the temperature of the body T(t).
(d) Determine an implicit expression for $t_d$, the time of death of the victim.

Relevant Equations

Integration

(a) $T_{a}(t)=t+5$

(b)
$$\frac{dT}{dt}=-k(T-T_{a}(t))$$
$$\frac{dT}{dt}=-k(T-t-5)$$
$$\frac{dT}{dt}=-kT+kt+5k$$

Is my working for (a) and (b) so far correct?

Thanks

 

[QuarkyMeson]

(a) Write down an expression for the atmospheric temperature Ta(t), where t = 0 at 1 pm.

So your (a) isn't fully done. What you wrote down is how the ambient temperature changes after 1pm.

(b) Write down an ODE to model the experiment. Solve this equation and use this to determine the value of k.

You're supposed to model the experiment. What is the ambient temperature in the experiment? How do you solve and determine k?

 

[songoku]

So your (a) isn't fully done. What you wrote down is how the ambient temperature changes after 1pm.

Sorry I don't understand. If I put $t = -7$, then I get $T_{a} (t)=-2^o C$ which is the ambient temperature at 6 am so I think the equation can also be used to find ambient temperature before 1 pm?

You're supposed to model the experiment. What is the ambient temperature in the experiment? How do you solve and determine k?

Is the ambient temperature constant? I thought it is not constant so I put my ambient temperature to be $t+5$, like my answer to (a)

I am planning to use integrating factor to solve the ODE but I am not sure whether my ODE is correct.

Thanks

 

[QuarkyMeson]

Sorry I don't understand. If I put $t = -7$, then I get $T_{a} (t)=-2^o C$ which is the ambient temperature at 6 am so I think the equation can also be used to find ambient temperature before 1 pm?


Is the ambient temperature constant? I thought it is not constant so I put my ambient temperature to be $t+5$, like my answer to (a)

I am planning to use integrating factor to solve the ODE but I am not sure whether my ODE is correct.

Thanks

Yeah my mistake, I read part (a) wrong. Your part (a) is correct.

For part (b) the experiment ambient temperature appears constant. The next clue would be that part (c) would just be a rehash of part (b) if they both evolved over time the same.

The pathologist determined from experiments that in a room at 5◦C the victim would cool from 15◦C to 12◦C in one hour.

 

[songoku]

For part (b) the experiment ambient temperature appears constant.

Oh ok.
$$\frac{dT}{dt}=-k(T-5)$$
$$\int \frac{1}{T-5}dT=\int -k~dt$$
$$\ln|T-5|=-kt+c$$

At $t=0, T=15$ and $t=1, T=12$ so I get $k=\ln \frac{7}{10}$ and $c=\ln 10$, so:
$$\ln |T-5|=\left(\ln \frac{7}{10}\right)t+\ln 10$$

The next clue would be that part (c) would just be a rehash of part (b) if they both evolved over time the same.

I don't understand part (c). I thought by "environment" the question means the ambient temperature so my answer for the ODE is $\frac{dT_a}{dt}=1$ and the solution is the same as part (a) which I don't think make sense at all.

 

[QuarkyMeson]

Oh ok.
$$\frac{dT}{dt}=-k(T-5)$$
$$\int \frac{1}{T-5}dT=\int -k~dt$$
$$\ln|T-5|=-kt+c$$

At $t=0, T=15$ and $t=1, T=12$ so I get $k=\ln \frac{7}{10}$ and $c=\ln 10$, so:
$$\ln |T-5|=\left(\ln \frac{7}{10}\right)t+\ln 10$$


I don't understand part (c). I thought by "environment" the question means the ambient temperature so my answer for the ODE is $\frac{dT_a}{dt}=1$ and the solution is the same as part (a) which I don't think make sense at all.

Your k is correct minus a sign getting lost.

It's probably easier to exponentiate everything to remove the logs. So $$\ln|T(t)-5|=-kt+c$$ becomes $$T(t) - 5 = Ce^{-kt}$$ or finally, $$ T(t) = 5+ Ce^{-kt}$$

In this case C = 10, while $k = -\ln \frac{7}{10}$ or $k = \ln \frac{10}{7}$

For part c you're now going to solve the differential equation you had before, namely $$\frac{dT}{dt}=-k(T-T_{a}(t))$$ With $T_{a}(t)$ equal to equation found in part (a).

 

[songoku]

Your k is correct minus a sign getting lost.

It's probably easier to exponentiate everything to remove the logs. So $$\ln|T(t)-5|=-kt+c$$ becomes $$T(t) - 5 = Ce^{-kt}$$ or finally, $$ T(t) = 5+ Ce^{-kt}$$

In this case C = 10, while $k = -\ln \frac{7}{10}$ or $k = \ln \frac{10}{7}$

Ah yes, I lost the negative sign.

For part c you're now going to solve the differential equation you had before, namely $$\frac{dT}{dt}=-k(T-T_{a}(t))$$ With $T_{a}(t)$ equal to equation found in part (a).

in part (c), is there only one question or are there two questions? I think "solve this equation" and "determine the temperature of the body T(t)" are same things.

Thanks

 

[QuarkyMeson]

You only need to find T(t) for part c.

 

[songoku]

Thank you very much QuarkyMeson

 

[QuarkyMeson]

Thank you very much QuarkyMeson

Anytime, let me know if you need any more help on parts (c) or (d).