Suppose a metal block is cooling so that its temperature $T$ (in ${ }^{\circ} \mathrm{C}$ ) is given by $\mathrm{T}=200 \cdot 4^{-0.1 \mathrm{t}}+20$, where $\mathrm{t}$ is in ho
a. Find the temperature after (i) 2 hours, (ii) 3.5 hours.
b. How long has the cooling been taking place if the block now has a temperature of $120^{\circ} \mathrm{C}$ ?
c. Find the eventual temperature $(t \rightarrow \infty)$.
a. (i) After 2 hours the temperature will be about $171.6^{\circ} \mathrm{C}$.
(Simplify your answer. Do not round until the final answer. Then round to the nearest tenth as needed.)
(ii) After 3.5 hours the temperature will be about $143.1^{\circ} \mathrm{C}$.
(Simplify your answer. Do not round until the final answer. Then round to the nearest tenth as needed.)
b. The cooling has been taking place about hours.
(Simplify your answer. Do not round until the final answer. Then round to the nearest integer as needed.)
Final Answer: \(\boxed{(i) 171.6^\circ \mathrm{C}, (ii) 143.1^\circ \mathrm{C}, 5 \text{ hours}, 20^\circ \mathrm{C}}\)
Step 1 :Given the temperature function \(T=200 \cdot 4^{-0.1t}+20\), where \(T\) is the temperature in degrees Celsius and \(t\) is the time in hours.
Step 2 :To find the temperature after a certain time, we substitute the time into the function.
Step 3 :(i) After 2 hours, the temperature is \(T=200 \cdot 4^{-0.1 \cdot 2}+20\), which simplifies to approximately \(171.6^\circ \mathrm{C}\).
Step 4 :(ii) After 3.5 hours, the temperature is \(T=200 \cdot 4^{-0.1 \cdot 3.5}+20\), which simplifies to approximately \(143.1^\circ \mathrm{C}\).
Step 5 :To find the time taken for the block to reach a certain temperature, we set the function equal to that temperature and solve for \(t\).
Step 6 :For a temperature of \(120^\circ \mathrm{C}\), we solve the equation \(120=200 \cdot 4^{-0.1t}+20\) to find \(t\), which is approximately 5 hours.
Step 7 :To find the eventual temperature as \(t\) approaches infinity, we take the limit of the function as \(t\) approaches infinity.
Step 8 :The limit as \(t\) approaches infinity of \(T=200 \cdot 4^{-0.1t}+20\) is \(20^\circ \mathrm{C}\).
Step 9 :Final Answer: \(\boxed{(i) 171.6^\circ \mathrm{C}, (ii) 143.1^\circ \mathrm{C}, 5 \text{ hours}, 20^\circ \mathrm{C}}\)