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D.E law of cooling

bergausstein

Active member
Jul 30, 2013
191
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.
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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

Active member
Jul 30, 2013
191
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?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,780
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.