A hot ballon in a small box placed in a room

From this question: from physics in everyday life, the temperature of the balloon is lowered to around $\approx 30^{\circ}$

What if, the balloon was placed in a box with equal empty volume left as of the balloon. Assuming the same conditions of box being $30^{\circ}$ and the balloon being $40^{\circ}$ . Then would the temperature of both be $\approx 35^{\circ}$
Also what if this entire system of box and balloon was placed in a room of $25^{\circ}$ . What would happen then? (Assuming the room has a far greater volume than the box)
Finally is there an empirical relation between volume and temperature. Like $PV=nRT$

#Mechanics

Easy Math Editor

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Comments

In the first condition, as the 📦 and 🎈 have almost the same volume of air, the temperature should be around 35 degree celsius, because when heat is exchanged it almost comes to a mean value and also as they have the same V, it would take some only some thermal energy from the 🎈 to heat up the box to get the same temp. However in the room scenario the room has a much greater volume so the system shouldn't heat the room air to a much higher degree as it would require a lot of heat to a raise the temperature by a very comparable extent, so the temperature might come almost near to 25 degree Celsius.

Siddharth Chakravarty - 11 months, 2 weeks ago

Thanks for answering. This is what I thought but I wasn't sure. Is there a formula or relation for volume and temperature??

Mahdi Raza - 11 months, 2 weeks ago

I would say the principle of heat exchange which states that the heat lost by one object is the same as the heat lost by the other object. Heat change is given by m x c x dt where m is the mass of the object, c is specific heat and dt is the change in temp. So if we use the principle of heat exchange for the scenarios, and calculate we find the temp. will be around as I stated by using a few assumptions like almost the same density, so that we can relate volume and etc. You can search the Web for more, or check this Brilliant course: Scientific essentials - Heat - Heat capacity.

Siddharth Chakravarty - 11 months, 2 weeks ago

@Siddharth Chakravarty – Also @Mahdi Raza PV=nRT is used to understand how change in temperature affects the volume but you are asking here how is temperature related to volume for heat exchange of 2 things of the same thing i.e air here. So principle of heat exchange has to be used.

Siddharth Chakravarty - 11 months, 2 weeks ago

@Siddharth Chakravarty – I can't find anything about heat exchange. Can you provide a link. Thanks!!

Mahdi Raza - 11 months, 2 weeks ago

@Mahdi Raza – Umm I mentioned you of the course, or you can search of the law of thermodynamics or heat capacity.

Siddharth Chakravarty - 11 months, 2 weeks ago

Hi Madhi, the idea that you're looking for is called heat capacity. You can find the relevant equation in this quiz.

In the situations that you're wondering about, energy is conserved, which means that if one part of the system gains energy $\Delta Q$ that energy needs to come from the other part of the system which loses $\Delta Q$ . The corresponding temperature changes depend on the mass and specific heats of the two parts of the system. You can calculate the exact $\Delta T$ by solving the system of equations of two instances of this equation such that the final $T$ is the same for both parts and $\Delta Q_1 + \Delta Q_2 = 0$ (or in other words, $\Delta Q_1 = - \Delta Q_2$ ).

Lee Weinstein Staff - 11 months, 2 weeks ago

Thanks @Lee Weinstein, The course I may say, is very well designed! Thank you for that!

Mahdi Raza - 11 months, 2 weeks ago

@Mahdi Raza, I said about the same concept and mentioned the same course before, feel free to ask if you want the calculations.

Siddharth Chakravarty - 11 months, 2 weeks ago

@Siddharth Chakravarty – Sure! Thanks to you as well!

Mahdi Raza - 11 months, 2 weeks ago

@Mahdi Raza Here is the calculation below:

We will assume the mass of the system from where the heat is being transferred from i.e the balloon as ${ m }_{ o }$ and the mass of the place where the heat is being transferred i.e the box or the room as ${ m }_{ i }$ . As the temperature range is small, and the heat is being transferred within the same substance air, the specific heat c could be assumed as same. Let the initial temperature of the ballon be ${ t }_{ o }$ , the initial temperature of the outside system as ${ t }_{ i }$ and the common temperature they reach after heat exchange is done as T. Thus the temperature difference for the ballon will be $\triangle { T }_{ o }$ = ${ t }_{ o }$ - T and for the outside system it will be $\triangle { T }_{ i }$ = T - ${ t }_{ i }$ .

We will now use the principle of heat exchange and formula for specific heat to calculate.

${ m }_{ o }c{ \triangle T }_{ o }={ m }_{ i }c{ \triangle T }_{ i }$

In the above equation c gets cancelled and we can expand $\triangle { T }_{ o }$ and $\triangle { T }_{ i }$ as the temperature differences stated above and we are left with the below equation:

${ m }_{ o }({ t }_{ o }-T)={ m }_{ i }(T-{ t }_{ i })$

Now we can substitute the mass in terms of density and volume as $Mass=Density \times Volume$ as the density will be the same for the same reason as the specific heat, which can be denoted as ρ and the volume with the respective subscripts, so substituting $m_{ o }=ρ { V }_{ o }$ and $m_{ i }=ρ { V }_{ i }$ . Now if ${ V }_{ i }$ is greater than ${ V }_{ o }$ k times it could be written as ${ V }_{ i }=k{ V }_{ o }$ and then substituting this for ${ m }_{ i }$ , we get ${ m }_{ i }$ = $ρk{ V }_{ o }$ and substituting now this new mass values in the above equation we get,

$ρV_{ o }({ t }_{ o }-T)={ ρkV }_{ o }(T-{ t }_{ i })$

Volume and density get cancelled, and then we rearrange the equation then so that T is on one side and the other terms on the other. We get,

$T=\frac { { t }_{ o }+k{ t }_{ i } }{ k+1 }$

We will will call the above equation we just derived as Mahdi's equation XD!

Now in the 1st scenarion, both the box and balloon have same volume so k=1, and solving the Mahdi equation with the respective values as ${ t }_{ o }$ = 40 degree C and ${ t }_{ i }$ = 30 degree C, we find T=35 degree C as you stated.

And in the second scenario, we could think of a big room where its volume is 1000 times that of a balloon so k = 1000 and ${ t }_{ o }$ = 35 degrees C and ${ t }_{ i }$ = 25 degrees C, we get the temperature almost as 25.009 which is almost as 25 degrees C.

Siddharth Chakravarty - 11 months, 1 week ago

Though it includes a lot of variables, nicely done! Did you derive that "Mahdi's equation" (sounds weird), or is that a standard rearrangement as well!

Mahdi Raza - 11 months, 1 week ago

I named it Mahdi's equation after your name. And yes I derived the Mahdi's equation which is useful when we are comparing two things of the same substance over a small temperature difference, the Mahdi's equation doesn't work for any other conditions though but The Principle of heat exchange is a general formula for any case.

Siddharth Chakravarty - 11 months, 1 week ago

Although this is a very good empirical relation between k, the volume, and given temperature t, can this be done for 3 bodies?

Mahdi Raza - 11 months, 1 week ago

Yes it could be done for n bodies but you need to properly account for the temperature changes and how to frame the equations accordingly as principle of heat exchange itself is just an application of conservation of energy. One example of 3 bodies is the calorimeter, with which we had to experiment in school.

Siddharth Chakravarty - 11 months, 1 week ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$