Question

1. a baker records the internal temperature of a pie that has been left to cool on a counter. the room temperature is 14°c. an equation that models this situation is (t(t)=68(0.5)^{\frac{t}{10}} + 14) where (t) is the temperature in degrees celsius and (t) is the time in minutes.

a) determine the temperature, to the nearest degree, of the pie after 30 minutes.
3a
b) how much time did it take for the pie to reach an internal temperature of 34°c?
3a

Explanation:

Step1: Substitute \(t = 30\) into the formula

Given \(T(t)=68(0.5)^{\frac{t}{10}}+14\), when \(t = 30\), we have \(T(30)=68(0.5)^{\frac{30}{10}}+14\).
First, calculate the exponent: \(\frac{30}{10}=3\). Then, \((0.5)^3=0.5\times0.5\times0.5 = 0.125\).
So \(T(30)=68\times0.125 + 14\).

Step2: Perform multiplication and addition

\(68\times0.125=8.5\). Then \(T(30)=8.5 + 14=22.5\approx23^{\circ}C\).

Step3: Solve for \(t\) when \(T(t)=34\)

Set \(T(t) = 34\), so \(34=68(0.5)^{\frac{t}{10}}+14\).
Subtract 14 from both sides: \(34 - 14=68(0.5)^{\frac{t}{10}}\), which gives \(20 = 68(0.5)^{\frac{t}{10}}\).
Then \((0.5)^{\frac{t}{10}}=\frac{20}{68}=\frac{5}{17}\).
Take the natural - logarithm of both sides: \(\ln((0.5)^{\frac{t}{10}})=\ln(\frac{5}{17})\).
Using the property of logarithms \(\ln(a^b)=b\ln(a)\), we get \(\frac{t}{10}\ln(0.5)=\ln(\frac{5}{17})\).
Since \(\ln(0.5)\approx - 0.693\), \(\ln(5)\approx1.609\), \(\ln(17)\approx2.833\), then \(\ln(\frac{5}{17})=\ln(5)-\ln(17)\approx1.609 - 2.833=-1.224\).
So \(\frac{t}{10}=\frac{\ln(\frac{5}{17})}{\ln(0.5)}=\frac{- 1.224}{-0.693}\approx1.77\).
Multiply both sides by 10 to find \(t\approx17.7\approx18\) minutes.

Answer:

a) \(23^{\circ}C\)
b) \(18\) minutes