D.E Law of Cooling: Find 9:20am Reading & Correct Temp

[bergausstein]

At 9am , a thermometer reading 70ºF is taken outdoors, where the temperature is 15ºF. At 9:05am, the thermometer reading is 45ºF. At 9:10am, the thermometer is taken back indoors, where the temperature is fixed at 70ºF.

(a). Find the reading at 9:20am

(b). when the reading, to the nearest degree, will show the correct (70ºF) indoor temperature.

I have already taken the specific solution using the conditions given in the problem,

$x(t)=55(\frac{11}{6})^{-\frac{t}{5}}+15$

plugging in $t=10$ I have $x=31.4 F$

now I don't know what to do next. please help!

and can you explain the question in b.

 

[I like Serena]

Hey bergausstein! :)

At 9am , a thermometer reading 70ºF is taken outdoors, where the temperature is 15ºF. At 9:05am, the thermometer reading is 45ºF. At 9:10am, the thermometer is taken back indoors, where the temperature is fixed at 70ºF.

(a). Find the reading at 9:20am

(b). when the reading, to the nearest degree, will show the correct (70ºF) indoor temperature.

I have already taken the specific solution using the conditions given in the problem,

$x(t)=55(\frac{11}{6})^{-\frac{t}{5}}+15$

plugging in $t=10$ I have $x=31.4 F$

Good!

now I don't know what to do next. please help!

Create a new formula with the same inherent parameters for the next phase.
Note that your current formula is of the form:
$$x(t)=\Delta T_0 (\frac{11}{6})^{-\frac{t}{5}}+T_\infty$$
where $\Delta T_0$ is the initial difference in temperature and $T_\infty$ is the final temperature.

and can you explain the question in b.

Theoretically the temperature will reach exactly 70ºF only after an infinite amount of time, so we're interested in when we would read it off as 70ºF.
To the nearest degree means that the temperature would be 69.5ºF.

 

[bergausstein]

the given in the new condition

let 9:10am as $t=0$ and 70F as my ambient temperature, and $\alpha=-\frac{\ln\frac{6}{11}}{5}$

now I'll have,

when $t=0$; $x(0)=31.4F$

then my new specific solution is

$x(t)=-38.64e^{(\frac{11}{6})^{\frac{-t}{5}}}+70$

the temperature at 9:20am is

since 9:10am is t=0, 9:20am is t=10

$x(10)=-38.64e^{(\frac{11}{6})^{\frac{-10}{5}}}+70=58.5F$

@9:20am $x=58.5ºF$

but I still don't get why 69.5 is the nearest degree?

 

[I like Serena]

the given in the new condition

let 9:10am as $t=0$ and 70F as my ambient temperature, and $\alpha=-\frac{\ln\frac{6}{11}}{5}$

now I'll have,

when $t=0$; $x(0)=31.4F$

then my new specific solution is

$x(t)=-38.64e^{(\frac{11}{6})^{\frac{-t}{5}}}+70$

the temperature at 9:20am is

since 9:10am is t=0, 9:20am is t=10

$x(10)=-38.64e^{(\frac{11}{6})^{\frac{-10}{5}}}+70=58.5F$

@9:20am $x=58.5ºF$

Looks good!

but I still don't get why 69.5 is the nearest degree?

Any actual temperature between 69.5ºF and 70.5ºF would be read as 70ºF, since we're reading the temperature "to the nearest degree".
Or put otherwise, we are reading the scale without any digits after the decimal point.

Since the temperature is rising, the first time this would be the case is when the temperature is 69.5ºF.

 

[blue_raver22]

Hello,

Thank you for sharing your solution to the problem. It looks like you have correctly calculated the temperature at 9:20am to be 31.4ºF.

For part (b), the question is asking at what time will the thermometer reading show the correct indoor temperature of 70ºF. To find this, we can set up the equation:

70 = 55(11/6)^(-t/5) + 15

Solving for t, we get t = 25 minutes. This means that at 9:25am, the thermometer reading will show the correct indoor temperature of 70ºF.

I hope this helps clarify the question for you. Keep up the good work in your scientific calculations!