a We will use the given function to model the situation.
T(t)= T_r+(T_i- T_r)e^(kt)
T(t) is the temperature of a heated substance t minutes after it has been removed from a heat or cooling source. T_r is room temperature, T_i is the substance's initial temperature, and k is a constant for that substance. In our case T_r= 72, T_i=236, and k=- 0.041.
T(t)= 72+(236- 72)e^(- 0.041t)
⇕
T(t)=72+164e^(- 0.041t)
We want to determine how long will it take for the roast to cool to 100^(∘)F. In other words, we want to determine for what t the value of T(t) is equal to 100.
100=72+164e^(- 0.041t)
Let's solve the above equation for t.
It will take about 43 minutes for the roast to cool to 100^(∘)F.
b In Part A we have found an equation that describes the temperature T of a heated substance t minutes after it has been removed from a heat or cooling source.
T=72+164e^(- 0.041t)
If we solve the above equation for t, we will be able to use it to check our answer from Part A and to complete the given table.
Now we can use a graphing calculator to graph the equation.
Next we will use the graph of the equation to check our answer from Part A and to complete the given table.
Checking the Answer
To check our answer form Part A, we have to check if t≈ 43 when T=100. This can be done by pressing 2nd, TRACE, and choosing the value option. Finally, enter 100 and it will give you the value of - 10.041ln( T-72164) when T= 100.
Rounded to the nearest integer, the value of - 10.041ln( 100-72164) is 43, so our answer from Part A is correct.
Completing the Table
Let's complete the given table. First, we want to determine the value of t when T= 225. Once more we will use the trace feature to find this value.
Rounded to one decimal place, the value of - 10.041ln( 225-72164) is 1.7. We will follow the same process for the remaining values.
