Question

What is the initial temperature of the liquid and when is the temperature at 100°F?T(m)=101e^-0.03m+67

Answer 1

a) The initial temperature of the liquid is simply the temperature of the liquid at time m = 0

Thus, we have the temperature to be:

`\begin{gathered} T(m)=101e^(-0.03m)+67 \\ \Rightarrow T(0)=101e^(-0.03*0)+67 \\ \Rightarrow T(0)=101e^0+67 \\ \Rightarrow T(0)=101(1)^{}+67 \\ \Rightarrow T(0)=101^{}+67 \\ \Rightarrow T(0)=168^oF \end{gathered}`

The initial temperature of the liquid is 168 degrees Fahrenheit

b) The time when the temperature of the liquid will reach 100 degrees Fahrenheit is obtained as follows:

`\begin{gathered} T(m)=101e^(-0.03m)+67 \\ \Rightarrow100=101e^(-0.03m)+67 \\ \Rightarrow100-67=101e^(-0.03m) \\ \Rightarrow33=101e^(-0.03m) \\ \Rightarrow(33)/(101)=e^(-0.03m) \\ \Rightarrow e^{\mleft\{-0.03m\mright\}}=(33)/(101) \end{gathered}`

Now, take the natural logarithm of both sides, as follows:

`\begin{gathered} e^{\mleft\{-0.03m\mright\}}=(33)/(101) \\ \Rightarrow\log _ee^{\{-0.03m\}}=\log _e((33)/(101)) \\ \Rightarrow(-0.03m)\log _ee=\log _e((33)/(101)) \\ \Rightarrow(-0.03m)(^{}1)=\log _e((33)/(101)) \\ \Rightarrow-0.03m=\log _e((33)/(101)) \\ \Rightarrow m=(1)/(-0.03)*\log _e((33)/(101)) \\ \Rightarrow m=(1)/(-0.03)*-1.1186=(-1.1186)/(-0.03)=37.39 \\ \Rightarrow m=37.4\text{ minutes (to the nearest tenth of a minute)} \end{gathered}`

The time when the temperature of the liquid will reach 100 degrees Fahrenheit is 37.4 minutes