[Bored chemist]

Sometimes the number you need is "about a hundred or a thousand" which is the right ballpark for how much denser a solid is, compared to the vapour near atmospheric pressure.

Which is why I suggested using a liquid. Or a sublimating solid.

Ok, because the vapour takes up an impractical amount of space you suggested turning a solid or liquid into the (impractical)  vapour.
Were you aiming to fail?

[Eternal Student]

Hi again.

I'll try and get it done but I'm off to the big city tomorrow and I know I won't be writing much here for at least a day.

    I'm really sorry but I'm not going to get this done any time soon.

   I'm already on pages of work.   Therefore no-one will want to read it, not even @chiralSPO .
I do not have the time to complete.  I'm way behind on several other important tasks.  The whole thing is now begining to cause me some stress and that's not how a forum should be. 
   I'm very sorry @chiralSPO  but you should not wait for a detailed response from me, it would safer to assume it won't happen.

Apologies,  Eternal Student.

[chiralSPO (OP)]

Thanks to Eternal Student and Bored chemist for trying to answer the questions I posed.

Hi again.

I'll try and get it done but I'm off to the big city tomorrow and I know I won't be writing much here for at least a day.

    I'm really sorry but I'm not going to get this done any time soon.

   I'm already on pages of work.   Therefore no-one will want to read it, not even @chiralSPO .
I do not have the time to complete.  I'm way behind on several other important tasks.  The whole thing is now begining to cause me some stress and that's not how a forum should be. 
   I'm very sorry @chiralSPO  but you should not wait for a detailed response from me, it would safer to assume it won't happen.

Apologies,  Eternal Student.

Thanks for trying. I do not mean to give anybody homework to stress over.

I too have gotten bogged down in the math, which is why I reached out.

I have gotten somewhere:

Assuming a two state model (A Z), the equilibrium ratio of concentrations is:

[Z]_T/[A]_T = e^{–RT/(ΔΗ–TΔS)}

Because we want ΔH–TΔS to be 0 at comfortable temperatures, that restricts the range of ΔH as a function of ΔS (or vice versa). The bigger ΔH is, the more thermal energy can be absorbed per mole of transformed molecule as T increases (and vice versa). But due to the exponential nature of the function providing [Z]/[A], there will be only a very narrow range in which there is an appreciable amount of both A and Z.

Because one is turning into the other, at a given temperature, T1, [Z]_T1 = 1–[A]_T1
K_T1= (1–[A]_T1)/[A]_T1
And then at a different temperature, T2, K_T2= (x+1–[A]_T1)/([A]_T1–x), where x is how many moles is getting converted.

amount of energy absorbed when going from T1 to T2 = ΔH•x

And then I get lost in the algebraic rearrangements when trying to get ΔH•x as a function of T1 and T2. (which then means I cannot use calc to optimize it).


I may keep banging my head against this wall for a while (hints or answers would still be appreciated). But I think the long and short of it is: there isn't really much advantage to trying for a gradual transition--it doesn't change the overall amount of energy that can be absorbed (ΔH). So, as Bored chemist says, maximizing ΔH/£ (or ΔH/$) for something with a comfortable equilibrium temperature is probably the way to go, and then just use as much as needed to keep the temp where you want it.

 

[Bored chemist]

It's worth remembering that, for decades, we used steam radiators The surface temperature of them was near 100 C.
But we used some sort of thermostat to maintain the temperature near 20C .
You can get away with the equilibrium temperature being a little "off" the temperature you want to maintain.

Having said that, using something that melts near 21C is a good starting point..
And if you have that, you automatically have the right value of delta S for the material's delta H.

So, as you say  what you want is something cheap  that "melts" near 21C and has a large delta H per $
I think soft paraffin wax is likely to be a good candidate.
The other thing you need to consider is thermal conductivity. Wax isn't so good at that.
 

[Eternal Student]

Hi again.

   @chiralSPO  - this is what I had started writing but stopped.   It's rough, not spell-checked and just not finished.   However, if it helps you,  then you can see it.     You'll realise why I stopped.... it's too long and I started by explaining things too simply, as if other users might also be reading it.

    Actually it just is too long....  the forum won't allow me to post anything with too many characters.....   I'll split it into two chunks.   I'll also put it under a "spoiler" so it won't clog up all the space for people who are reading this thread.
- - -  - - - - - - - - -

Chunk 1

Spoiler: show

But what I am most interested in at the moment, is just the math involved in figuring out how to maximize the ability of the solution to absorb/release energy over a desired temperature range.

    Let's minimise peoples disinterest and just do something for the simple situation.
   
You have the simple chemical reaction forming an equilibrium   A Z

   Since the reaction is 1 molecule  to/from  1 other molecule,  we have

N_A   + N_Z  =  N   = a constant at all times 

[Eqn 1]

   with  N_A = number of A molecules;   N_Z = no. of Z molecules;   N = total number of molecules.

We're going to tackle this situation with the equilibrium constant.  I've considered alternative strategies but this one is reasonably easy, makes good progress and produces answers that seem to agree with what you might have expected.

K  =   [Z] / [A]   =  ratio of concentrations at equilibrium.   

[Eqn 2]

      The way you've described it we can assume the reactants and products are in the same container,  possibly in some solvent....  Whatever details apply - they slosh around in or contribute to some fluid of volume V.   The volume might change slightly as products and reactants come and go, or temperature causes expansion and contraction, that won't matter.  All that matters is that at any instant, the products and reactants are in the same volume.   So we get some cancellation and can replace concentrations with numbers of molecules in the equilibrium constant.

     Hence,    K  =  N_Z  / N_A.     

[Eqn 3]

Combine [Eqn 1]  and [Eqn 3]  to obtain,

    N_A  =     

[Eqn 4]

That's a decent start,  we know the number of A molecules that should be present at any time.   Note that the equilibrium constant K  is a not a constant, it changes depending on various parameters.

   The balance of the reaction equilibrium will be entirely determined by K.   We take this expression for K
   

[Eqn 5]

A reference for this result was given in a previous post.   Here's a slightly different one:  https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/Free_Energy_and_Equilibrium
     This is where we will make contact with your (@chiralSPO) ideas of varying the enthalpy and entropy.  We haven't forgotten about it.  All the information and contact we shall need will come through the equilibrium constant.  That constant tells us how the equilibrium shifts. 
     We're going to need a moment to explain this formula [Eqn 5].  As already mentioned in an earlier post, some texts would have used ΔG° in place of ΔG in that formula and everyone has a slightly different view on what ΔG° is supposed to be.   In the worst case, some texts insist that ΔG° has been determined at a standard reference temperature, in that situation ΔG° is nothing more than a fixed value you can find in a book, a constant regardless of whatever else is happening.
    Now we MUST HAVE a quantity,  ΔG,   which depends on temperature.   Consider for a moment that it doesn't but instead we are using a quantity which I'll call ΔG" and which is just a constant regardless of whatever else is happening.  An equilibrium with K described by [Eqn 3] achieves a perfect balance with an equal amount of product and reactant   if and only if     K = 1.     Next use [Eqn 5] with ΔG = ΔG" = a fixed constant:
    K =  e ^(-ΔG"/RT).   

[Eqn 6]

So   K = 1   if and only if    one of the following holds:   (1)  Temperature T →∞  ,  or else  (2) ΔG" = 0.   We're not interested in infinite temperatures, so we would want ΔG" = 0.   However ΔG" was assumed to be a constant, so using  [Eqn 6] again,  we have   K = e⁰ =  1   regardless of temperature.    The equilibrium never changes, it doesn't respond to temperature at all.   We've discussed this before - it's no use.
   The approximation where a fixed standard value ΔG" was used in [Eqn 5] is acceptable for many situations especially where you are working around standard temperatures AND  ΔG" is not approximately 0.   However, that is not our situation.  We are very likely to want an equilibrium operating around the perfect balance point with K ≈ 1 and hence ΔG" ≈ 0.    We must have a better approximation for the equilibrium constant.
 
   Here's [Eqn 5] again:

[Eqn 5]

That's a perfectly good expression where ΔG  is allowed to be determined by temperature.   Many texts will still call the value that will be inserted into that formula ΔG° but it is fully assumed to be dependant on temperature.  It is just the free energy change that occurs for the forward reaction  A → Z  at a given temperature.   (Further discussion of this removed, this post is already too long).
Anyway, this ΔG is calculated as follows:
     ΔG =  ΔH - TΔS

[Eqn 7]

   As @chiralSPO  mentioned,   ΔH   and  ΔS  do have some temperature dependence but over the range of temperatures we are considering it's not important.   ΔH ≈ ΔH° = constant   and  ΔS ≈ ΔS° = constant.     We obtain,
    ΔG°  =   ΔG°(T) =     ΔH° - TΔS°

[Eqn 8]

    Note that this ΔG°  does retain some temperature dependence and it is sufficient for our purposes,   the quantity T is a variable and appears as the multiple of ΔS°.   (It's only the more extreme definitions of ΔG° where the temperature would have been locked at one value and become what I previously described as ΔG" ).

   I appreciate we might lose some people here.  It would help to get a diagram to show what we're doing:

Sketch to show  ΔG°   and  K  as functions of T  given by Eqn 8 and Eqn 5

graph2.png (19.6 kB . 1495x935 - viewed 2578 times)

Features of the sketch:    K = 0 at T=0 kelvin.      K  → finite value = e^ΔS°/R  as T →∞.   It has a point of inflection around  T=ΔH° / 2R.     
   ΔG°  is just a straight line,  with  ΔG° = 0  at  temp. T₀ = ΔH°/ΔS°   (easily obtained from Eqn 8 ).
   The point of inflection for K(T) is not too important but is easily obtained from the expression   K = e^-ΔG°/RT  =  (e^ΔS°/R) .  e^-ΔH°/RT    =    (constant) . e^-(ΔH°/RT)   and differentiating twice with respect to  T.   
     The most important characteristic is that K passes through 1  at T = T₀  this is where the equilibrium would have an equal amount of product and reactant.

---End of Chunk 1 ---

[Eternal Student]

Chunk 2

Spoiler: show

So the question becomes. For a given high and low temperatures (T_high and T_low), what is the optimal choice of ΔH and ΔS, such that the system is most stable between those to temperatures?

    Stability is not something you've decleared very well.   I'm going to interpret the notion of "stability" as implying that the reaction would absorb (or release) as much heat as possible to counter a temperature change in the room.
   Let's just spend a moment to see that this is a sensible approach:

Diagram 2:   How the room loses / gains thermal energy

   Thermal Energy change
   in the Room,  cΔT               ----> heat lost to the outside world, ΔE   (due to... poor insulation etc.)
       /\
        |
        | 
  Heat released by chemical equillibrium system, ΔU


    Under the situation shown in the above diagram we have     
Change in thermal energy of the room  =   c.ΔT  =    ΔU - ΔE                 

[eqn 10]

  with c = heat capacity of the room;   ΔE = energy transferred out of the room;   ΔU = heat released by the chemical system into the room.
   Note the sign conventions we have established:    Room cooling: ΔT<0   ;  ΔU >0 the chemical system will release heat (thermal energy) into the room  ;  ΔE > 0  heat is lost to the outside world.    Let's not spend too long discussing this, it should be apparent that ΔE can be considered as the sum of natural heat exchange to the world outside the room + energy exchanged due to heating or cooling systems like radiators or air-conditioning units that a person has inside the room.  A negative value indicates a net heat input to the room (e.g. from the house central heating system).
    Note also that the chemical system is considered as part of the contents of the room.   In the most degenerate situation the room is only the chemical system but usually the room is the chemical system + the rest of the room (like walls and furniture) such that the chemical system is just a minor contribution to the mass and heat capacity for the room.   (LATE EDITING:  Stuff removed - the above is just conservation of energy ideas and doesn't need further comment).

   The following results follow from [Eqn 10]
(i) ΔU ≠ ΔE  iff the room changes temperature.
(ii)  .. result removed in late editing... serious error here, sorry.   The following result still holds but for different reasons...
   Therefore,  |ΔU| ≤ |ΔE| always and equality holds if and only if ΔT =0.   

[Result 11]

   We can now see that no chemcal equillibrium system A↔Z   exists which would perfectly maintain a fixed temperature in the room.    Assume some energy ΔE≠0 is removed from (or added to) the room.  For ΔT = 0, we must have ΔU = ΔE  but  the chemical equillibrium would not change at all if the temperature didn't change, so we would have ΔU = 0.    Therefore, when some energy enters or leaves the room we are always in the situation with |ΔU| < |ΔE| and the best we can hope for is that the chemical equillibrium opposes  ΔE as best it can (so that the temperature change will be less severe than if the chemical equillibrium system wasn't there).

    I would suggest that the best interpretation of "maximum stability" for our temperature buffering system will be trying to keep |ΔT| as low as possible for a given amount of ΔE (energy transferred to/from something outside the room).  For our purposes, the quantity ΔE is beyond our control or influence, that requires someone to insulate the room and turning on/off the central heating system of the house etc.

    From [Eqn 10] ; [Results 11] and recognising that under our sign conventions ΔU opposes the sign of ΔT we have,
   |ΔT| = 

[Equation 12]

So, that minimising the size of the temperature change for the room due to a given Energy transfer into / out of the room, requires the size of |ΔU| to be MAXIMISED.

     CLAIM  that ΔU = heat released from the chemical system = function of the current Temperature T, the parameters ΔH° and ΔS° of the reaction A→B   and, of course, of the temperature change ΔT that occurrs.  PROOF: We have the following,
   ΔU =  ΔN_A . ΔH° / (Avogadro)   

[Equation 13]

 
   Where, ΔN_A = the change in N_A (change in the number of molecules of A);     ΔH° is the specific enthalpy of the reaction change (heat change for 1 mol of A being converted to Z);     (Avogadro) = Avogadros number.   The signs work (I hope),  recall ΔN_A is negative, N_A decreases, when molecules change from A to Z   and  ΔH° = negative when  A→Z is exothermic,  making  ΔU = heat released to the room = positive.
    This is taking too long.... let's just speed this up.... we can obtain an expression for ΔN_A   from  [Equation 4] in terms of the change in the equillibrium constant K.   We have a thermodynamic expression for K from [equation 5].   That establishes the "CLAIM" that ΔU is a function of the things mentioned above.  The dependance on ΔT (the change in temperature) came from ΔN_A but that will be clearer in a moment...

    During the actual operation of the chemical equillibrium device,  you cannot change the parameters ΔH° or ΔS°  of the reaction or add extra molecules of A,  or do very much to change the way the device can abosrb or release heat.   The ony changes that occurr are temperature changes in the room.  So in calculating the quantity ΔN_A during the actual operation of the device,   we can treat all other parameters as fixed constants.   This saves a mess writing derivatives as partial derivatives,  for example  K = K(T) = a function only of Temperature during actual operation,   even though  K has dependance on ΔH and ΔS  looking at [eqn 5].   This should also help to explain why finding partial derivatives w.r.t.  variables like  ΔH°   is not something we will do.   Just to be clear on this:  It might seem that since we are trying to maximise ΔU  (amount of heat released by the chemical system) with a carefull selection of parameters like ΔH° for the reaction, then we would want to find derivatives w.r.t. ΔH° and treat the problem as a typical maximisation problem.   However, that is not what we will do.   Instead, we treat the reaction parameters as a fixed constant (don't specify the actual value - just recognise that it is a constant) and see how the system will naturally behave in operation.   After that, we are free to see how those results change or what they will mean when the reaction parameters like ΔH° and ΔS° are given a particular numerical value.   This method seems to work better than just directly trying to find maxima from derivatives w.r.t.  reaction parameters.   (For example, there doesn't seem to be any local maximum.   This might be discussed later -  the larger ΔH° can be the better, the progression is endless and no local maxima exists  BUT also  you just don't get much useful information about the system anyway).

   Let's see [Eqn 13] again:
δU =  δN_A . ΔH° / (Avogadro)   

[Equation 13' ]

     We'll drop the Avogadros number (as if ΔH° was stated per molecule not per mole), it only scales the value of ΔU,  the maximum won't be affected without it anyway.   Note that we are now considering only small changes, as marked with δ symbols -   the quantity  ΔH°  remains with a big delta because it is just a reaction parameter, a constant and not something that changes during the operation of the device.  We'll also just write ΔH instead of ΔH°  because it's shorter and should be clear enough that this is a fixed constant, a reaction parameter not something that changes dynamically during operation.
   From [Eqn 4] we can differentiate, noting that temperature is the only variable that changes during operation of the device:
   δN_A   =   
K and the derivative dK/dT can be found from [Eqn 5] and [Eqn 7]
    = 

   

[Check the signs work here:  If ΔH >0,  then A→Z is endothermic.  K = [Z]/[A]  and we expect K to increase when T increases.  dK/dT >0  is what we have.  Next up δN_A has a negative sign out at the front, so it's negative and yes we have A→Z so the number of molecules of A would decrease.   This all looks OK to me. ]

  Substitute this into [Eqn 13' ] and we have:

δU =      

[Eqn 14]

   where  ζ =  e^ΔS/R  =  a convenient constant to hold all the entropy information.   Note that a lot of the quantities appearing in {Eqn 14] are constants,    T is the only variable.

Recall that we wish to maximise δU  given a fixed transfer of energy δE from the room.  Straight away from [Eqn 14]  we can see that  if N (the total number of molecules of A and Z) is increased then δU is increased.   There is also a term on the numerator  ~  (ΔH)²   and although ΔH does appear elsewhere its effect is most noticeable here. *LATE EDITING:  Reasonably true for small to medium ΔH.   However, when ΔH becomes very large, then the exponential function e^-ΔH will dominate and bring the numerator towards 0.   This effect is more important than I first thought because ΔH usually is large - hundreds of thousands in standard units of Joules/mol. * 
   We can tidy up our expression a little.   We are trying to maximise δU  for a fixed value δE of energy transferred out of (or into) the room.   However, we can imagine the maximisation requirement differently.   Assume that a small change in temperature (δT) of the room has happened.   We would like to have this cause a release (or absorption) of energy by the chemical system  (δU)  that is as large as possible because that would have forced δE to be as large as possible.   [Eqn 12]  can be used to formalise this condition and establish that it is equivalent.  The post is too long, I'm skipping that.  We will just re-state the maximisation problem as required:   To buffer the temperature of the room as much as possible,  we wish to maximise δU  for a given fixed change in temperature δT.    Anyway, with that modification we can divide by δT  and obtain the following result (where δT→0 to form the derivative on the R.H.S.)

   

[Eqn 15]

     Our objective is to maximise this quantity.

.....  I stopped writing here.... there are some derivatives and some graphical plots to show what is happening that i would have inlcuded.... but by now   no-one is reading this   and the whole set of expressions have become sufficiently messy,   no-one wants to see them anymore.   They belong on a computer system like SymbolLab or Mathematica... or whatever your preferred mathematical manipulation software might be.   They do not belong on a text document for people to read or differentiate by hand.

    Minor Notes:    We were really maximising  |δU|  ,  the size of a change.  As it happens  δU  will oppose  δT in sign,   which means that we are often actually trying to minimise  δU  because  its negative and a lower negative has a greater size.   I started modifying everything to |δU|  in the early part of the post... but time limits cut in... I never carried it through to the end of the post.
    Here are some plots of what we get from [Eqn 15].

Parameters   N=1  (arbitrary scalar anyway).         ΔH = 10 000   J/mol       ΔS = 33.33  J/mol/K    (deliberately chosen so that  ΔH/ΔS ≈ 300 kelvin ≈ approx. room temperature  and ΔG would become 0 at that temp).

Plot-300.JPG (66.42 kB . 1363x418 - viewed 2617 times)

   Note that although I would have expected the minima   (maxima of |dU/dT| ) to be  around  T = 300 k,  it was actually a bit below that.


Paremeters as before,   ΔH raised to 20 000 J/mol.    ΔS = 33.33.     Expected  T = 600 kelvin but the maxima is still slightly off.

plot-600.JPG (82.3 kB . 1401x535 - viewed 2606 times)


Anyway, the general  behaviour of the chemical system seems to be reasonable.....   it always has maximum temperature buffering ability at a particular value of temperature.   Changing the values of ΔH  and  ΔS   move that  maxima   up or down.     You can probably also see why you can't get clear results by taking something like [Eqn 15] and directly differentiating w.r.t.  reaction parameters like ΔH.....      You can have a system that is actually tuned to buffer around a temperature of  100 kelvin *   instead of  300 kelvin   BUT   if  the value of N (the number of molecules in the system) is large enough and also  ΔH   is large eneough then it would still be a numerically better buffer at room temperature than a smaller and more sensibly ΔH valued system.    * LATE EDITING in the previous sentence:  It's more noticeable where the system is tuned to buffer below room temperature not above.  See the actual plots with given parameters.  It is, in my opinion, only by analysing the situation with ΔH and ΔS as fixed parameters and recognising that T is the only thing that can chnage in operation of the device where the natural behaviour of the device is observed.   For a realistic situation, where you can't have an infintely sized system or infinitely high values of ΔH,  you would want to work with the natural behaviour of the system and set it up with ΔH and ΔS  parameters that put it's maximum buffering ability right at the room temperature you wish to try and maintain.

Best Wishes.

[chiralSPO (OP)]

thank you! it will take some time to read, but it looks like you made more headway than I did, so this should be very informative!

[Bored chemist]

You may find this useful.
https://en.wikipedia.org/wiki/Trouton%27s_rule

[Eternal Student]

Hi.

   Come on then @chiralSPO  let's have an update on how you're doing.

[Z]T/[A]T = e–^{RT/(ΔΗ–TΔS)}

    Presumably you noticed that you had the exponent upside down.     e^{-(ΔH-TΔS)/RT}    should have been there.

   However, overall the approach using  the equilibrium constant, K, is what we both went for.

    I think I kept mentioning that ΔH should be as large as possible  BUT   it's actually a bit indirect....  it's the size of ΔS that makes the more obvious difference  however you need  ΔH  to be increasing with it just to keep the maximum response of the chemical equilibrium in the right sort of position (at the right temperature).

    Anyway... by now I was thinking you might have looked at the last formula,  I think it was labelled [Eqn 15] and decided if I made some errors along the way to getting there.   If it's alright, you can continue with it, that's what I would have done.   That's a thing you can examine directly and optimise it by varying the ΔH and ΔS parameters of the reaction.

   Anyway, I don't think you need your two temperatures  T₁   and T₂    just a single figure  T  where you hope to keep the temperature of the room.

   You can develop the idea and see what happens with more complicated reactions involving multiple particle species and having an equilibrium constant involving products of concentrations raised to some powers.  I'm not saying I want to do it.... the algebra is only going to get more messy... just that it could be done.

Best Wishes.

[Bored chemist]

  I think I kept mentioning that ΔH should be as large as possible  BUT   it's actually a bit indirect....  it's the size of ΔS that makes the more obvious difference 

You are talking as if they are independent properties, but once you decide on the temperature you want, there's not much choice.
At equilibrium delta G is zero,
so T Delta S = delta H

What you want to do is store heat and that's delta H so you want that to be big (for a given mass, volume, price or whatever).

[Eternal Student]

Hi.

You are talking as if they are independent properties, but once you decide on the temperature you want, there's not much choice.

    I completely agree.   That was the essence of it. 

    There is one, if quite silly, choice you can make that does allow you to have ΔS and ΔH values chosen independently.    You can develop the chemical equilibrium system to have ΔG of the reaction = 0 at some other temperature (not the desired room temperature) and just accept the consequences.   You'd have the equilibrium constant k >> 1   or  else  k <<1   but never the less,  it would be some equilibrium with  0 < k < ∞.   By Le Chatelier's principle alone if you want to keep it simple, the system will oppose a change in temperature.  It can't perform as well as when the system had k≈1 but it will function to some extent.   So you can compensate by just making the system massive,  i.e. have many molecules of the stuff in the reaction.   
    To phrase this another way, a massive but badly set-up equilibrium system can still oppose temperature changes and help to hold a room at a given temperature.  If it is massive enough then it will be better than a smaller system set up to have ΔG ≈ 0 right at approx. room temperature.
    Other than cost of production, the massive but badly set up system may have other undesirable properties:   It may not help the person in the room to establish the preferred room temperature as much as the finely tuned and well set up system would.   It just maintains it once you have got the room to the preferred temperature initially.   Specifically, the system becomes less responsive on one side of the preferred temperature than on the other side of that temperature.  Let's say it was set up with ΔH and ΔS values to be an optimal temperature buffer at 0 deg. C then it's effectiveness falls off as you move further away from 0 deg C.   So that the person can push the room temperature above the preferred temperature by accident just by leaving their heating system on for an extra half-hour.  Meanwhile, if the equilibrium system is set up to have maximum temperature bufferring ability precisely at 22 deg C then it's ability to absorb / release additional heat remains more consistent on both sides of the preferred temperature.  Therefore it's more forgiving of a careless person who left the heating on for an extra half hour by mistake - they won't raise the room temperature that much higher than the preferred value.

Best Wishes.