Newton's Law of cooling & Exponential growth

[wind]

Hi, can someone check my work? Thanks

A cup of hot chocolate, in a room temperature of 21c, cools according to Newton's lae of cooling. Determine the rate of cooling, k, of the hot chocolate if it cools from 86c to 65c in 15min.

\(\displaystyle \L\ T-T_{s}=(T_{o}-T_{S})e^{kt}\)

\(\displaystyle \L\ 65-21=(86-21)e^{k15}\)

\(\displaystyle \L\ 44=(65)e^{k15}\)

\(\displaystyle \L\ log44=log(65)e^{k15}\)

\(\displaystyle \L\ log44=k15log(65)e\)

\(\displaystyle \L\frac{log44}{(log65)(e)(15)}=k\)

\(\displaystyle \L\ 0.022=k\)


2. A population of termites is increasing according to the formula
\(\displaystyle \L\ P=p_{0}e^{kt}\). Determine the lenght of time, t, that it atkes theis population to triple its initial population of 1800 if it doubles in 0.35 days.


\(\displaystyle \L\ P=p_{0}e^{kt}\)

\(\displaystyle \L\ 5400=1800e^{0.035t}\)

\(\displaystyle \L\ log5400=log1800e^{0.035t}\)

\(\displaystyle \L\ log5400=(0.035t)(log1800e)\)

\(\displaystyle \L\frac{log5400}{(log1800e)(0.035)}=t\)

\(\displaystyle \L\28.9=t\)

3. A van's engine has overheated to 190c, so the driver pulls over to the side of the road and shuts off the engin. The engin cools to 150c in 5min. The engin must cool to 80c before the driver can start the van again. If the outside temperature if 28c, how long will it be before the driver can restart the van?

\(\displaystyle \L\ T-T_{s}=(T_{o}-T_{S})e^{kt}\)

\(\displaystyle \L\ 80-28=(190-28)e^{kt}\)

..I don't know how to do this one


Thanks

 

[galactus]

3. A van's engine has overheated to 190c, so the driver pulls over to the side of the road and shuts off the engin. The engin cools to 150c in 5min. The engin must cool to 80c before the driver can start the van again. If the outside temperature if 28c, how long will it be before the driver can restart the van?

\(\displaystyle \L\ T-T_{s}=(T_{o}-T_{S})e^{kt}\)

\(\displaystyle \L\ 80-28=(190-28)e^{kt}\)

..I don't know how to do this one


Thanks

This isn't algebra. It's more suited to the DE forum.

Use your data to find k and then solve.

\(\displaystyle \L\\\frac{dT}{dt}=k(T-28), \;\ T(0)=190\)

Separate variables and integrate:

\(\displaystyle \L\\\int\frac{dT}{T-28}=\int{kdt}\)

\(\displaystyle \L\\ln|T-28|=kt+C_{1}\)

\(\displaystyle T=28+C_{2}e^{kt}\)

When t=0, T=190. Therefore,

\(\displaystyle 190=28+C_{2}\)

\(\displaystyle C_{2}=162\)

Hence, \(\displaystyle T=28+162e^{kt}\)

Now, use your given data to find k and you're all set.

 

[wind]

This isn't algebra. It's more suited to the DE forum.

Differential Equations? I'm in grade 12 calculus. Maby there is a simpiler way to do this?

Find K

T=150
Ts=28
To= 190
K=?
t=5

\(\displaystyle \L\ T-T_{s}=(t_{o}-T{s})e^{kt}\)

\(\displaystyle \L\ 150-28=(190-28)e^{k5}\)

\(\displaystyle \L\ 122=(162)e^{k5}\)

\(\displaystyle \L\ 0.7530=e^{k5}\)

\(\displaystyle \L\ ln0.7530=k5lne\)

\(\displaystyle \L\ -0.0567380102=k\)

the plug in

T=80
Ts=28
To= 190
K=-0.0567380102
t=?

would that work?

btw I did not know how to put degree signs so i just put "c", sorry if that caused any confusion...

and for the others i used "log" instead of "ln" right?