Question

A body at an unknown temperature is placed in a room which is held at a constant temperature of $30^{\circ}F$. If after 10 minutes the temperature of the body is $0^{\circ}F$ and after 20 minutes the temperature of the body is $15^{\circ}F$, then the expression for the temperature of the body at any time $t$ is

• A. $T = -60 e^{-0.069 t} - 30$
• B. $T = -60 e^{-0.03010 t} + 30$
• C. $T = 60 e^{-0.069 t} + 30$
• D. $T = 60 e^{-0.069 t} - 30$

Hint:

For cooling problems, if the temperature approaches an ambient value, the form is $T(t) = T_{ambient} + (T_0 - T_{ambient})e^{-kt}$.

Answer 1

The Correct Option is B

Step 1: Set up Newton's cooling model.
The body cools toward the room temperature of $30$. The standard model is $T(t)=30+Ce^{-kt}$, where $C$ and $k$ are constants we must find.
Step 2: Use the reading at 10 minutes.
At $t=10$, $T=0$, so $0=30+Ce^{-10k}$, which gives $Ce^{-10k}=-30$.
Step 3: Use the reading at 20 minutes.
At $t=20$, $T=15$, so $15=30+Ce^{-20k}$, which gives $Ce^{-20k}=-15$.
Step 4: Divide to remove C.
Dividing the second by the first: $e^{-10k}=\tfrac{-15}{-30}=\tfrac{1}{2}$. So the value $e^{-10k}=0.5$.
Step 5: Find C.
From $Ce^{-10k}=-30$ and $e^{-10k}=0.5$, we get $C(0.5)=-30$, so $C=-60$.
Step 6: Write the temperature law.
Putting the constants back, $T=30-60e^{-kt}$, i.e. $T=-60e^{-kt}+30$, which matches the form in option (B).
\[ \boxed{\,T=-60e^{-0.03010\,t}+30\,} \]