Question

a large room has a temperature of 70 degrees fahrenheit. when a cup of tea is brought into the room, the tea has a temperature of 120 degrees fahrenheit. the function f(t)=ce^(-kt)+70 represents the situation, where t is time in minutes, c is a constant, and k is a constant. after 3 minutes the cup of tea has a temperature of 100 degrees. what will be the temperature of the tea, in degrees fahrenheit, after 5 minutes? round your answer to the nearest tenth, and do not include units. provide your answer below:

Explanation:

Step1: Find the value of $C$

When $t = 0$, $f(0)=120$. Substitute into $f(t)=Ce^{-kt}+70$. We get $120 = C e^{0}+70$. Since $e^{0}=1$, then $120=C + 70$, so $C=120 - 70=50$. The function becomes $f(t)=50e^{-kt}+70$.

Step2: Find the value of $k$

When $t = 3$, $f(3)=100$. Substitute into $f(t)=50e^{-kt}+70$. We have $100=50e^{-3k}+70$. First, subtract 70 from both sides: $100 - 70=50e^{-3k}$, so $30 = 50e^{-3k}$. Then $e^{-3k}=\frac{30}{50}=0.6$. Take the natural - logarithm of both sides: $\ln(e^{-3k})=\ln(0.6)$. Using the property $\ln(e^{x})=x$, we get $-3k=\ln(0.6)$, and $k=-\frac{\ln(0.6)}{3}\approx-\frac{- 0.5108}{3}\approx0.1703$.

Step3: Find the temperature at $t = 5$

Substitute $C = 50$, $k\approx0.1703$, and $t = 5$ into $f(t)=50e^{-kt}+70$. So $f(5)=50e^{-0.1703\times5}+70$. First, calculate $-0.1703\times5=-0.8515$. Then $e^{-0.8515}\approx0.4267$. Multiply by 50: $50\times0.4267 = 21.335$. Add 70: $f(5)=21.335+70=91.335\approx91.3$.

Answer:

91.3