# Thread: chemical reaction and newton's cooling

1. ## chemical reaction and newton's cooling

Thanks Jhevon,
Yeah, the normal body temperature is 37. I just asked my teacher about it.

another two practical problems would have help me a lot. Moving to chemical problem.

1. A compound C is formed when two chemicals A & B are combined. The resulting reaction between the two is such that for each gram of B, 3 grams of A are used. Initially there are 36 grams of A and 18 grams of B. It's observed that 10 grams of the compound C are formed in 15 minutes. Determine the amount of C at any time if the rate of the reaction is proportional to the amounts of A & B remaining. Find the amount after 20 minutes.

2. Another newton's cooling problem.

On Easter, a ham is removed from an oven and placed in a room where the temperature is 70 degree F. A meat thermometer indicates that internal temperature of the ham to be 220 degree F. 30 minutes later, the meat thermometer indicates 200 degree F. How much longer it will take to cool the meat to 100 degree F.

3. Grow and Decay

The wood of an Egyptian sarcophagus (burial case) is found to contain 63% of the Carbon-14 found in a present-day sample. What is the age of the sarcophagus if the half-life of C14 is 5730 years ?

again, thanks Jhevon....I appriciate that.

2. Originally Posted by ggw
1. A compound C is formed when two chemicals A & B are combined. The resulting reaction between the two is such that for each gram of B, 3 grams of A are used. Initially there are 36 grams of A and 18 grams of B. It's observed that 10 grams of the compound C are formed in 15 minutes. Determine the amount of C at any time if the rate of the reaction is proportional to the amounts of A & B remaining. Find the amount after 20 minutes.
i'll have to think about this problem some more. there's one part of it that is persistent in confusing me.

2. Another newton's cooling problem.

On Easter, a ham is removed from an oven and placed in a room where the temperature is 70 degree F. A meat thermometer indicates that internal temperature of the ham to be 220 degree F. 30 minutes later, the meat thermometer indicates 200 degree F. How much longer it will take to cool the meat to 100 degree F.
Ok, you know the drill here. we will use the formula:

$T' = -k(T - T_{amb})$

$\Rightarrow T = Ae^{-kt} + T_{amb}$ ...see my previous posts on these types of questions to see how I derived this.

$\Rightarrow T = Ae^{-kt} + 70$

Now $T(0) = 220$

$\Rightarrow 220 = Ae^0 + 70$

$\Rightarrow A = 150$

Also, $T(30) = 200$

$\Rightarrow 200 = 150e^{-30k} + 70$

$\Rightarrow k = \frac {\ln \left( \frac {13}{15} \right)}{-30}$

$\Rightarrow k \approx 0.00477$

So our equaiton is: $T = 150e^{-0.00477t} + 70$

Now just plug in $T = 100$ and solve for $t$

When done, you have to subtract 30 from that value, since the question asked how much LONGER it will take to cool to 100 [after 30 mins].

3. Grow and Decay
For growth and decay, we use pretty much one formula. The only difference is, in decay, we use a negative exponent. As a mnemonic, I call this the PERT formula, as in the shampoo Pert, if you remember--that was a great invention, saves a lot of time. Anyway, what does PERT mean? See if you can see the word below:

$P = P_0 e^{rt}$

Notice that the left side of the formula spells PERT. In the above formula, $P$ (a function of time) is the amount of substance after time $t$, $P_0$ is the initial amount of substance, $r$ is the rate of growth, and $t$ is the time elapsed. This is known as the formula for Exponential Growth, and is usually used to predict the growth of bacteria and populations.

For the Exponential Decay formula, we simply make the exponent of $e$ negative. So for exponential decay:

$P = P_0 e^{-rt}$

Note that some books will use different letters for $P$. Other common letters used are $A \mbox{ and } I$.

one more formula you need to know for this topic is:

$r \cdot t_h = \ln 2$
where $r$ is the rate of decay, and $t_h$ is the half-life.

The wood of an Egyptian sarcophagus (burial case) is found to contain 63% of the Carbon-14 found in a present-day sample. What is the age of the sarcophagus if the half-life of C14 is 5730 years ?
Now, $r \cdot t_h = \ln 2$

$\Rightarrow r = \frac {\ln 2 }{t_h} = \frac { \ln 2}{5730} \approx 0.000121$

This s a decay problem, so we use the formula:

$A = A_0 e^{-rt}$

So we have: $A = A_0 e^{-0.000121t}$

Now we are not told the amount we have now, or the amount that we started with, so how do we proceed here? Answer: we pick a number. I like to use 1, but any number you pick will work out fine.

So assume we start with a sample containing 1 unit Carbon-14. We now have 63% of that, so now we have 0.63 units of Carbon-14. So we can let $A_0 = 1$, so now we just have to find $t$ such that, $A = 0.63$. Thus, to answer this question, we must solve:

$0.63 = e^{-0.000121t}$

for $t$.

$t$ will give you the age of the sample

the formula for this is dX/dt = k(amount A remaining)(amount B remaining)
for each:

2 parts of A & 3 parts of B are being used.

Original: 36 of A, 18 of B

10 grams are formed in 15 minutes.

X = 10

A = ? , B = ?

dX/dt = k(36 - A)(18 - B)

So 2x/5 for A, B would be 3x/5.

Determined the amount of C at any time if the rate of the reaction is proportional tothe amounts of A & B Remaining. Find the amount after 20 minutes.

4. Originally Posted by ggw
the formula for this is dX/dt = k(amount A remaining)(amount B remaining)
for each:

2 parts of A & 3 parts of B are being used.

Original: 36 of A, 18 of B

10 grams are formed in 15 minutes.

X = 10

A = ? , B = ?

dX/dt = k(36 - A)(18 - B)

So 2x/5 for A, B would be 3x/5.

Determined the amount of C at any time if the rate of the reaction is proportional tothe amounts of A & B Remaining. Find the amount after 20 minutes.

you're using X here to mean C? i thought it was 3 grams of A uses 1 gram of B to form C.

anyway. let's see if the blind can lead the blind through this. going with the 2 parts A and 3 parts B.

if 10 g of C is formed in 15 mins, 2/3 g is formed in one minute. 2/5 of that is from A, and 3/5 of that is B. So after every minute, we use (2/5)(2/3) = 4/15 g A and (3/5)(2/3) = 2/5 g B.

then, $\frac {dC}{dt} = k \left(36 - \frac {4}{15}t \right) \left(18 - \frac {2}{5} t\right)$

$\Rightarrow \frac {dC}{dt} = k \left( 648 - \frac {96}{5} t + \frac {8}{75} t^2 \right)$

$\Rightarrow dC = k \left( 648 - \frac {96}{5}t + \frac {8}{75} t^2\right) dt$

$\Rightarrow \int dC = k \int \left( 648 - \frac {96}{5}t + \frac {8}{75} t^2 \right) dt$

$\Rightarrow C = k \left(648t - \frac {48}{5} t^2 + \frac {8}{225} t^3\right)$

Now when $t=15$, $C = 10$:

use this to find $k$. then you can use that equation to find $C$ when $t = 20$