Why is my answer to this ap calc question completely completely off? :( help please! Thank you!!

Question

anonymous · Answer

what is the question? what is your answer?

anonymous · Answer

oops Ill attached the photo

anonymous · Answer

Do you see the question now?

anonymous · Answer

There's an error in your fourth line:
$$200=-\frac{6.89}{0.053}e^0+C~~\implies~~200=-\frac{6.89}{0.053}+C~~\implies~~C=330$$

anonymous · Answer

thank you!!!

anonymous · Answer

But then I get 224 for my answer and its not right?

anonymous · Answer

Oh I didn't notice the other mistake you made earlier. When you integrated to find $T(t)$, you have two negatives that should cancel out:
$$-6.89\int e^{-0.053t}\,dt=\frac{-6.89}{-0.053}e^{-0.053t}+C=\color{red}{130}e^{-0.053t}+C$$

anonymous · Answer

ohh right thank you! and so C comes out to be 70, and is that why the question tells you to room temperature is 70 degrees on average? Because I thought that was just a trick in the question as we are dealing with tea temp not room temp..but why would that just be the constant?

anonymous · Answer

Well as the tea cools indefinitely, it will reach the ambient temperature. Notice that as $t\to\infty$, the exponential term approaches $0$, so you're left with $70^\circ$, the room temperature.