A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is
. After one minute the thermometer reads .
(a) What will the reading on the thermometer be after 4 more minutes?
(b) When will the thermometer read ?
minutes after it was taken to the outdoors.

Ask by Kirk Fernandez. in the United States

Mar 22,2025

Question

A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is
. After one minute the thermometer reads .
(a) What will the reading on the thermometer be after 4 more minutes?
(b) When will the thermometer read ?
minutes after it was taken to the outdoors.

Ask by Kirk Fernandez. in the United States

Mar 22,2025

UpStudy AI · Accepted Answer

We start with Newton's law of cooling. The temperature of the thermometer at time $$ t $$ is given by

$$
T(t) = T_{\text{env}} + \bigl(T(0)-T_{\text{env}}\bigr)e^{-kt},
$$

where  
- $$ T_{\text{env}} $$ is the ambient temperature,
- $$ T(0) $$ is the initial temperature of the thermometer,
- $$ k $$ is a positive constant.

In this problem:
- The ambient (outdoors) temperature is $$ T_{\text{env}} = -13^\circ \mathrm{C} $$.
- The initial reading was taken indoors, so $$ T(0) = 19^\circ \mathrm{C} $$.

Thus, the general formula becomes

$$
T(t) = -13 + \left(19 - (-13)\right)e^{-kt} = -13 + 32e^{-kt}.
$$

At $$ t = 1 $$ minute, the thermometer reads $$ 8^\circ \mathrm{C} $$. Substituting this into our model:

$$
8 = -13 + 32e^{-k}.
$$

Solve for $$ e^{-k} $$:

$$
8 + 13 = 32e^{-k} \quad \Longrightarrow \quad 21 = 32e^{-k},
$$

$$
e^{-k} = \frac{21}{32}.
$$

Taking the natural logarithm of both sides, we have

$$
-k = \ln\left(\frac{21}{32}\right) \quad \Longrightarrow \quad k = -\ln\left(\frac{21}{32}\right) = \ln\left(\frac{32}{21}\right).
$$

---

**(a) Temperature after 4 More Minutes (i.e. $$ t = 5 $$ minutes):**

Using the formula with $$ t = 5 $$ minutes:

$$
T(5) = -13 + 32e^{-5k}.
$$

Since $$ e^{-k} = \frac{21}{32} $$, we have

$$
e^{-5k} = \left(e^{-k}\right)^5 = \left(\frac{21}{32}\right)^5.
$$

Thus,

$$
T(5) = -13 + 32\left(\frac{21}{32}\right)^5.
$$

This is the exact expression for the temperature after 5 minutes.

For an approximate numerical value, compute:

- First, 
  $$
  \frac{21}{32} \approx 0.65625.
  $$
- Then, 
  $$
  (0.65625)^5 \approx 0.121.
  $$
- Finally,
  $$
  32 \times 0.121 \approx 3.872.
  $$

So,

$$
T(5) \approx -13 + 3.872 \approx -9.13^\circ \mathrm{C}.
$$

---

**(b) Time When the Thermometer Reads $$ -12^\circ \mathrm{C} $$:**

We set

$$
T(t) = -12,
$$

so

$$
-12 = -13 + 32e^{-kt}.
$$

Solve for $$ e^{-kt} $$:

$$
-12 + 13 = 32e^{-kt} \quad \Longrightarrow \quad 1 = 32e^{-kt},
$$

$$
e^{-kt} = \frac{1}{32}.
$$

Taking the natural logarithm:

$$
-kt = \ln\left(\frac{1}{32}\right) = -\ln(32),
$$

which gives

$$
t = \frac{\ln(32)}{k}.
$$

Recall that $$ k = \ln\left(\frac{32}{21}\right) $$. Hence, the time can be written as

$$
t = \frac{\ln 32}{\ln\left(\frac{32}{21}\right)}.
$$

For an approximate numerical value, compute:

- $$ \ln 32 \approx 3.4657 $$,
- $$ \ln\left(\frac{32}{21}\right) \approx \ln(1.5238) \approx 0.422 $$.

Thus,

$$
t \approx \frac{3.4657}{0.422} \approx 8.21 \text{ minutes}.
$$

---

$$
\textbf{Answers:}
$$

$$
\text{(a) } T(5) = -13 + 32\left(\frac{21}{32}\right)^5 \approx -9.1^\circ \mathrm{C} \quad (\text{5 minutes after being taken outdoors}).
$$

$$
\text{(b) } t = \frac{\ln 32}{\ln\left(\frac{32}{21}\right)} \approx 8.2 \text{ minutes after being taken outdoors}.
$$
After 5 minutes, the thermometer reads approximately $$ -9.1^\circ \mathrm{C} $$. It will reach $$ -12^\circ \mathrm{C} $$ in about 8.2 minutes after being taken outdoors.

A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is
. After one minute the thermometer reads .
(a) What will the reading on the thermometer be after 4 more minutes?
(b) When will the thermometer read ?
minutes after it was taken to the outdoors.

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A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is . After one minute the thermometer reads . (a) What will the reading on the thermometer be after 4 more minutes? (b) When will the thermometer read ? minutes after it was taken to the outdoors.

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Latest Physics Questions

A thermometer is taken from a room where the temperature is to the outdoors, where the temperature is
. After one minute the thermometer reads .
(a) What will the reading on the thermometer be after 4 more minutes?
(b) When will the thermometer read ?
minutes after it was taken to the outdoors.