Question

A thermometer is removed from a room where the temperature is 70 F and is t temperature is 30° F. After one-half minute the thermometer reads 50° F. What is the reading of the thermometer at and is taken outside, where the air t1 min? (Round your answer to two decimal places.) How long will it take for the thermometer to reach 35° F? (Round your answer to two decimal places.) min

Answer 1

It will take approximately 1.69 minutes (rounded to two decimal places) for the thermometer to reach 35°F when taken outside.

To solve this problem, we can use Newton's Law of Cooling, which describes how the temperature of an object changes when it is removed from one environment (in this case, a room) and exposed to a different environment (outside). The law is given by the formula:

`\[T(t) = T_e + (T_0 - T_e) \cdot e^(-kt)\]`

Where:

`\(T(t)\)` is the temperature of the object at time t.

`\(T_e\)` is the temperature of the environment (outside).

`\(T_0\)` is the initial temperature of the object when it was removed from the room.

k is a constant that depends on the object and the environment.

Given:

`\(T_0\)` (initial temperature) = 70°F

`\(T_e\)` (outside temperature) = 30°F

- After 30 seconds (t = 0.5 minutes), the thermometer reads 50°F.

We need to find the value of k first using the information at t = 0.5 minutes:

`\[50 = 30 + (70 - 30) \cdot e^(-0.5k)\]`

Let's solve for k:

`\[20 = 40 \cdot e^(-0.5k)\]`

Now, divide both sides by 40:

`\[0.5 = e^(-0.5k)\]`

Take the natural logarithm (ln) of both sides to isolate k:

`\[\ln(0.5) = \ln(e^(-0.5k))\]`

Using the property of logarithms
`\(\ln(e^x) = x\):`

`\[\ln(0.5) = -0.5k\]`

Now, solve for k:

`\[k = -(\ln(0.5))/(0.5)\]`

Calculate k:

`\[k ≈ 1.3863\]`

Now that we have the value of k, we can use it to find when the thermometer reaches 35°F:

`\[35 = 30 + (70 - 30) \cdot e^(-kt)\]`

Substitute the known values:

`\[5 = 40 \cdot e^(-1.3863t)\]`

Divide both sides by 40:

`\[0.125 = e^(-1.3863t)\]`

Take the natural logarithm (ln) of both sides:

`\[\ln(0.125) = \ln(e^(-1.3863t))\]`

Using the property of logarithms
`\(\ln(e^x) = x\`:

`\[\ln(0.125) = -1.3863t\]`

Now, solve for t:

`\[t = (\ln(0.125))/(-1.3863)\]`

Calculate t:

`\[t ≈ 1.6931\]`