A potato with temperature 70 degrees F is placed into a oven with an internal temperature 400 degree after 30 minutes, the temperature of the potato has increased to 205 degrees F.

Give the heating constant k (rounded to 6 decimal places), and find an expression for the temperature potato as a function of time.

Question

A potato with temperature 70 degrees F is placed into a oven with an internal temperature 400 degree after 30 minutes, the temperature of the potato has increased to 205 degrees F.

Give the heating constant k (rounded to 6 decimal places), and find an expression for the temperature potato as a function of time.

nottim · Answer

so useful...
whoa...was gunna give the anseer there for some reason...

anonymous · Answer

huh?

nottim · Answer

well, what's our initial constant? What would the entire thing equal?

anonymous · Answer

A potato with temperature 70 degrees F is placed into a oven with an internal temperature 400 degree. After 30 minutes, the temperature of the potato has increased to 205 degrees F.

nottim · Answer

uh

anonymous · Answer

You need the Newton's formula for this one.$$T(t)=T_1+(T_0-T_1)e ^{-kt}$$T0 is the starting temperature of the object, T1 is the temperature of the environment.

anonymous · Answer

As t gets larger, the second term fades away, which leads to the temperature of the object T(t) being T1 (the temperature of the surroundings.

anonymous · Answer

So plug the data from the problem into the formula; you will have only one unknown value (k) that can be solved for.

anonymous · Answer

ok

anonymous · Answer

Solve this....$$205=400+(70-400)e^{-30k}$$

anonymous · Answer

what is the temperature of the potato after 80 minutes?

anonymous · Answer

After what amount of time will the temperature of the potato reach 350 degrees F?

anonymous · Answer

You can find that after you have solved for k.

anonymous · Answer

The last unknown is k; once you have that, you just plug the time or temperature into the formula, and there will be only one unknown to solve for.

anonymous · Answer

$$205=400+(70−400)e^{−30k}\implies \frac{-195}{-330}=e^{−30k}$$$$\implies \frac{\ln(\frac{195}{330})}{-30}=k$$

anonymous · Answer

After you have k (use your calculator to get an approximation), you can now solve for unknown time or temperature easily.

anonymous · Answer

so 80 replaces 205 in the equation to solve for the temperature after 80 minute?

anonymous · Answer

No, you put 80 in for t, the time in the exponent.  T(t) is the temperature of the object as a function of t (time).  Previously, the time was 30 in the problem above.  See it?