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A thermometer reading [tex]$62^{\circ} F$[/tex] is placed in an oven preheated to a constant temperature. Through a glass window in the oven door, an observer records that the thermometer reads [tex]$232^{\circ} F$[/tex] after [tex]$\frac{1}{2}$[/tex] minute and [tex]$308^{\circ} F$[/tex] after 1 minute.

How hot is the oven?

Round the answer to the nearest multiple of 10.

[tex]\square[/tex] [tex]{}^{\circ} F[/tex]

Answer :

Let the temperature of the thermometer at time [tex]$t$[/tex] (in minutes) be given by Newton’s law of heating:

[tex]$$
T(t) = T_{\text{oven}} + \bigl(T(0) - T_{\text{oven}}\bigr) e^{-kt},
$$[/tex]

where
- [tex]$T(0)=62^\circ F$[/tex] is the initial thermometer reading,
- [tex]$T_{\text{oven}}$[/tex] is the constant temperature inside the oven, and
- [tex]$k$[/tex] is a positive constant.

We are given:

- After [tex]$0.5$[/tex] minute:
[tex]$$
T(0.5)=232^\circ F,
$$[/tex]
- After [tex]$1$[/tex] minute:
[tex]$$
T(1)=308^\circ F.
$$[/tex]

Thus, we obtain the equations

[tex]$$
232 = T_{\text{oven}} + (62 - T_{\text{oven}}) e^{-0.5k} \quad \text{(1)}
$$[/tex]

[tex]$$
308 = T_{\text{oven}} + (62 - T_{\text{oven}}) e^{-k} \quad \text{(2)}
$$[/tex]

To simplify the notation, let

[tex]$$
E = e^{-0.5k}.
$$[/tex]

Then note that

[tex]$$
e^{-k} = E^2.
$$[/tex]

Now the equations become

[tex]$$
232 = T_{\text{oven}} + (62 - T_{\text{oven}})E \quad \text{(1')}
$$[/tex]

[tex]$$
308 = T_{\text{oven}} + (62 - T_{\text{oven}})E^2 \quad \text{(2')}
$$[/tex]

### Step 1. Eliminate [tex]$T_{\text{oven}}$[/tex]

First, solve equation (1') for [tex]$T_{\text{oven}}$[/tex]. Rearranging (1') gives:

[tex]$$
232 = T_{\text{oven}} + (62 - T_{\text{oven}})E.
$$[/tex]

Write it as:

[tex]$$
232 = T_{\text{oven}} \,(1 - E) + 62E.
$$[/tex]

Then, we can solve for [tex]$T_{\text{oven}}$[/tex]:

[tex]$$
T_{\text{oven}} (1 - E) = 232 - 62E
$$[/tex]

[tex]$$
T_{\text{oven}} = \frac{232 - 62E}{1 - E}. \quad \text{(3)}
$$[/tex]

### Step 2. Find [tex]$E$[/tex]

Substitute [tex]$T_{\text{oven}}$[/tex] from (3) into equation (2'). Equation (2') is

[tex]$$
308 = T_{\text{oven}} + (62 - T_{\text{oven}})E^2.
$$[/tex]

Rather than substitute directly (which can get messy), observe an alternative method: Multiply equation (1') by [tex]$(1+E)$[/tex]:

[tex]$$
(232 - 62E)(1+E) = T_{\text{oven}}(1+E) + (62 - T_{\text{oven}})E(1+E).
$$[/tex]

However, a simpler elimination is possible by noticing that if we rewrite equations (1') and (2') in the form of rearranged expressions, we can eliminate [tex]$T_{\text{oven}}$[/tex]. Multiply (1') by [tex]$(1+E)$[/tex] so that its left side becomes

[tex]$$
(232 - 62E)(1+E) = 232(1+E) - 62E(1+E).
$$[/tex]

Expanding gives

[tex]$$
232 + 232E - 62E - 62E^2 = 232 + 170E - 62E^2.
$$[/tex]

Now set this equal to the right side, which according to (2') simplifies to

[tex]$$
308 - 62E^2.
$$[/tex]

Thus, we have:

[tex]$$
232 + 170E - 62E^2 = 308 - 62E^2.
$$[/tex]

Cancel the [tex]$-62E^2$[/tex] terms on both sides:

[tex]$$
232 + 170E = 308.
$$[/tex]

Solve for [tex]$E$[/tex]:

[tex]$$
170E = 308 - 232 = 76,
$$[/tex]

[tex]$$
E = \frac{76}{170} = \frac{38}{85} \approx 0.4471.
$$[/tex]

### Step 3. Determine the Constant [tex]$k$[/tex]

Recall that

[tex]$$
E = e^{-0.5k}.
$$[/tex]

Taking the natural logarithm of both sides:

[tex]$$
-0.5k = \ln(E),
$$[/tex]

so

[tex]$$
k = -2 \ln(E).
$$[/tex]

Substituting the value of [tex]$E$[/tex]:

[tex]$$
k \approx -2 \ln\left(\frac{38}{85}\right) \approx 1.6101.
$$[/tex]

### Step 4. Calculate the Oven Temperature [tex]$T_{\text{oven}}$[/tex]

Using equation (3):

[tex]$$
T_{\text{oven}} = \frac{232 - 62E}{1 - E}.
$$[/tex]

Substitute [tex]$E \approx 0.4471$[/tex]:

- Compute the numerator:

[tex]$$
232 - 62(0.4471) \approx 232 - 27.73 \approx 204.27.
$$[/tex]

- Compute the denominator:

[tex]$$
1 - 0.4471 \approx 0.5529.
$$[/tex]

Thus,

[tex]$$
T_{\text{oven}} \approx \frac{204.27}{0.5529} \approx 369.45^\circ F.
$$[/tex]

### Step 5. Round the Answer

Rounding [tex]$369.45^\circ F$[/tex] to the nearest multiple of [tex]$10^\circ F$[/tex], we obtain:

[tex]$$
T_{\text{oven}} \approx 370^\circ F.
$$[/tex]

### Final Answer

The oven is approximately [tex]$$370^\circ F.$$[/tex]

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