Radioactive decay & Newton's Method of Cooling

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lamaclass

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I needed help with one of both types of these problems.

1. The rise of decomposition of a gas is proportional to the amount of gas present. If 250 ml of gas decomposes to leave only 200 ml in 30 minutes, how long will it take for 100 ml of the original 250 ml to decompose?

Here's the work I did for it but I am not sure if I have the correct answer:

y=Ce[sup:3gl2cklz]kt[/sup:3gl2cklz]

200=250e[sup:3gl2cklz]30k[/sup:3gl2cklz]

4/5=e[sup:3gl2cklz]30k[/sup:3gl2cklz]

ln(4/5)/30 = k

100=250e[sup:3gl2cklz]ln(4/5)/30t[/sup:3gl2cklz]

2/5=e[sup:3gl2cklz]ln(4/5)/30t[/sup:3gl2cklz]

ln (2/5)/ln(4/5)/30=.1369 and then multiplied that into 30 minutes to get roughly 4.1 minutes as my final answer.

2.Newton's law of cooling states that the rate at which the temperature of an object changes is proportional to the differences in the temperature of the object and the temperature of the surronding environment. A metal plate that has been heated cools from 180 degrees to 160 degrees in 30 minutes when surronded by air at a temperature of 60 degrees. Use Newton's Law of Cooling to approximate its temperature at the end of one hour of cooling.

I'm not sure how to go about setting this one up. :?
 
2.Newton's law of cooling states that the rate at which the temperature of an object changes is proportional to the differences in the temperature of the object and the temperature of the surronding environment. A metal plate that has been heated cools from 180 degrees to 160 degrees in 30 minutes when surronded by air at a temperature of 60 degrees. Use Newton's Law of Cooling to approximate its temperature at the end of one hour of cooling.

I'm not sure how to go about setting this one up. :?
Click to expand...


Newton's cooling law \(\displaystyle \frac{dT}{dt}=k(T-60)\), where 60 is the room temperature.

The initial conditions, \(\displaystyle T(0)=180, \;\ T(\frac{1}{2})=160\) allow us to find C and k.

We can separate variables:

\(\displaystyle \frac{DT}{T-60}=kdt\)

Integrate:

\(\displaystyle ln(T-60)=kt+C\)

e to both sides:

\(\displaystyle T-60=e^{kt+C}\Rightarrow T-60=e^{kt}\cdot e^{C}\)

But \(\displaystyle e^{C}\) is a constant we can rename \(\displaystyle C_{1}\), and get:

\(\displaystyle T-60=C_{1}e^{kt}\)

Now, use the initial conditions to find k and C1. After that, all you need to do is plug in t=1 to find the temp at 1 hour.
 
\(\displaystyle First \ one:\)

\(\displaystyle You're \ good \ until \ the \ end.\)

\(\displaystyle 150 \ = \ 250\bigg(\frac{4}{5}\bigg)^{t/30}, \ not \ 100 \ = \ 250\bigg(\frac{4}{5}\bigg)^{t/30}, \ Why?\)

\(\displaystyle Note: \ that's \ \frac{t}{30}, \ not \ 30t.\)
 
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