A frozen cake at \(-18\,^\circ\text{C}\) is placed in a kitchen at \(22\,^\circ\text{C}\). The temperature difference between the cake and the room decreases exponentially. After \(10\) minutes, the difference is \(32\,^\circ\text{C}\).
a) Find the initial temperature difference.
b) Find the decay factor per minute for the temperature difference, and write a function for the difference after \(t\) minutes. Round the per-minute decay factor to four decimal places.
c) Find the cake's temperature after \(30\) minutes.
d) Make a table of the temperature difference after \(0\), \(20\), \(40\), and \(60\) minutes. Round each table value to the nearest hundredth.
Hints
- Account for the negative initial temperature when finding the difference.
- Use the value after \(10\) minutes to find either a 10-minute factor or a 1-minute factor.
- The cake approaches room temperature from below.
- For the table, use the fact that the difference is multiplied by \(0.8\) every \(10\) minutes.
Solution
1. The initial difference is \(22-(-18)=40\,^\circ\text{C}\).
2. If \(a\) is the decay factor per minute, then \(40a^{10}=32\). Thus \(a^{10}=0.8\), so \(a=0.8^{1/10}\approx0.9779\). The exact model is \(D(t)=40\left(0.8^{1/10}\right)^t\).
3. After \(30\) minutes, \(D(30)=40\cdot(0.8)^3=20.48\,^\circ\text{C}\). Since the cake is below room temperature, its temperature is \(22-20.48=1.52\,^\circ\text{C}\).
4. \(D(0)=40.00\), \(D(20)=25.60\), \(D(40)=16.384\approx16.38\), and \(D(60)=10.48576\approx10.49\), all in degrees Celsius.
Answer
a) \(40\,^\circ\text{C}\)
b) \(a=0.8^{1/10}\approx0.9779\); \(D(t)=40\left(0.8^{1/10}\right)^t\)
c) \(1.52\,^\circ\text{C}\)
d)
<table>
<tr>
<th>Time (min)</th>
<th>Temperature difference (°C)</th>
</tr>
<tr>
<td>\(0\)</td>
<td>\(40.00\)</td>
</tr>
<tr>
<td>\(20\)</td>
<td>\(25.60\)</td>
</tr>
<tr>
<td>\(40\)</td>
<td>\(16.38\)</td>
</tr>
<tr>
<td>\(60\)</td>
<td>\(10.49\)</td>
</tr>
</table>