Math  /  Calculus

QuestionFor the following exercises, use this scenario: A pot of hot soup with an internal temperature of 103103^{\circ} Fahrenheit was taken off the stove to cool in a 64Froom64^{\circ} \mathrm{Froom}. After fifteen minutes, the internal temperature of the soup was 98F98^{\circ} \mathrm{F}. Use Newton's Law of Cooling to write a formula that models this situation. Round to 5 decimal places if necessary T(t)=T(t)= \square To the nearest minute, how long will it take the soup to cool to 80F80^{\circ} \mathrm{F} ? \square minutes.
To the nearest degree, what will the temperature be after 2 and a half hours? \square F{ }^{\circ} \mathrm{F}

Studdy Solution

STEP 1

What is this asking? We've got some hot soup cooling down, and we need to figure out how long it takes to reach a certain temperature using Newton's Law of Cooling, and also what the temperature will be after a specific time. Watch out! Don't mix up the starting temperature of the soup with the room temperature!
Also, keep track of your units – we're dealing with minutes and hours, so conversions might be necessary.

STEP 2

1. Define Newton's Law of Cooling
2. Find the constant kk
3. Determine the time to cool to 80°F
4. Calculate the temperature after 2.5 hours

STEP 3

Newton's Law of Cooling tells us how things cool down over time.
The formula is T(t)=A+(T0A)ektT(t) = A + (T_0 - A)e^{-kt}, where T(t)T(t) is the temperature at time tt, AA is the surrounding temperature (our room temperature), T0T_0 is the **initial temperature** of the soup, and kk is a constant we need to figure out.

STEP 4

Let's **plug in** what we know!
The room temperature, AA, is 64F\mathbf{64^\circ F}.
The initial temperature of the soup, T0T_0, is 103F\mathbf{103^\circ F}.
So, our formula becomes T(t)=64+(10364)ektT(t) = 64 + (103 - 64)e^{-kt}, which simplifies to T(t)=64+39ektT(t) = 64 + 39e^{-kt}.

STEP 5

We know that after fifteen minutes (t=15t = \mathbf{15}), the soup's temperature is 98F\mathbf{98^\circ F}.
Let's **plug** these values into our formula: 98=64+39e15k98 = 64 + 39e^{-15k}.

STEP 6

Now, let's **solve for** k\mathbf{k}!
First, subtract 64 from both sides: 9864=39e15k98 - 64 = 39e^{-15k}, which gives us 34=39e15k34 = 39e^{-15k}.

STEP 7

Next, **divide** both sides by 39: 3439=e15k\frac{34}{39} = e^{-15k}.

STEP 8

To get rid of the exponential, take the natural logarithm (ln) of both sides: ln(3439)=15k\ln(\frac{34}{39}) = -15k.

STEP 9

Finally, **divide** by -15 to isolate kk: k=ln(3439)150.00887k = \frac{\ln(\frac{34}{39})}{-15} \approx \mathbf{0.00887}.
Our formula is now T(t)=64+39e0.00887tT(t) = 64 + 39e^{-0.00887t}.

STEP 10

We want to find the time tt when the temperature T(t)T(t) is 80F\mathbf{80^\circ F}.
So, we set up the equation: 80=64+39e0.00887t80 = 64 + 39e^{-0.00887t}.

STEP 11

Subtract 64 from both sides: 8064=39e0.00887t80 - 64 = 39e^{-0.00887t}, giving us 16=39e0.00887t16 = 39e^{-0.00887t}.

STEP 12

Divide by 39: 1639=e0.00887t\frac{16}{39} = e^{-0.00887t}.

STEP 13

Take the natural logarithm of both sides: ln(1639)=0.00887t\ln(\frac{16}{39}) = -0.00887t.

STEP 14

Finally, solve for tt: t=ln(1639)0.0088797t = \frac{\ln(\frac{16}{39})}{-0.00887} \approx \mathbf{97} minutes.

STEP 15

We need to find T(t)T(t) when tt is 2.5 hours.
Since our formula uses minutes, let's convert 2.5 hours to minutes: 2.5 hours60 minutes1 hour=1502.5 \text{ hours} \cdot \frac{60 \text{ minutes}}{1 \text{ hour}} = \mathbf{150} minutes.

STEP 16

Now, plug t=150t = 150 into our formula: T(150)=64+39e0.00887150T(150) = 64 + 39e^{-0.00887 \cdot 150}.

STEP 17

Calculate the exponent: 0.00887150=1.3305-0.00887 \cdot 150 = -1.3305.

STEP 18

Calculate the exponential: e1.33050.2643e^{-1.3305} \approx 0.2643.

STEP 19

Finally, calculate the temperature: T(150)=64+390.264364+10.3174FT(150) = 64 + 39 \cdot 0.2643 \approx 64 + 10.31 \approx \mathbf{74^\circ F}.

STEP 20

T(t)=64+39e0.00887tT(t) = 64 + 39e^{-0.00887t}.
It will take approximately **97** minutes for the soup to cool to 80F80^\circ \mathrm{F}.
After 2 and a half hours, the temperature will be approximately **74**°F\mathrm{°F}.

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