Gchem

[Glycogen]

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There is a Q in Destroyer that says:

A compound with a very small specific heat will give rise to a large deltaT.

I choose this as being a false statement but apparently it is true!!!

My logic: We define specific heat as the measure of the heat energy required to increase the temperature of an object by a certain temperature interval.
So, in this case when we have a small S.H. our delta T also has to decrease since the difference between [final-initial}T will be low!

I might be wrong,but would you explain this to me.
To be more specific,this is Q#123 in old version Destroyer.
Thanks.

 

[osimsDDS]

Equation is this: deltaH or q=mc deltaT

So since c goes down then deltaT must increase to keep the q constant...you can also do this with m and deltaT or m and c...

It's like the gas qualities: PV=k if you increase P by a number V has to decrease by that number...

 

[Dmitry Malayev]

The answer is correct, it is true, here is why mathematically the formula goes as such:

Delta H = (m)(c)(Delta T)
Delta H = Heat
m = mass
c = specific heat
Delta T = Change in Temperature
If we solve for c we get:
c= (Delta H)/((m)(Delta T))

This shows us that c and Delta T (Temp Change) are inversly related to each other. Meaning the larger T gets, the smaller the c will be.

And you can test this.
Assums m=1 and Delta H=50 in our case. Now use a very small Delta T and your c will be huge, but if you use a very large Delta T, your c will be extremely small. Therefore we can conclude that the smaller specific heat is, the greater the Change in Temperature will be.

Hope this helps! 🙂

 

[Glycogen]

Thank you very much.It truly did help.
Ok,since you guys are fast,I have another Q #120 of destroyer!
Maybe it is the old version or maybe it is my calculation but I think when we calculate net energy released of a reaction we normally multiply number of bonds by bond energy but in this Q,it has not.I was wonder why.
If you don't have the same Q,I can happily type it here.
Thanks again.

 

[TexasDent]

I don't have destroyer but the formula you would use to calculate this is deltaE = Break - Make.
So, you calculate the energy of all of the reactant bonds and subtract the energy of all of the bonds in the products made.
Which will give you the amount of energy the reaction releases.

I'm not sure if it is deltaE or deltaH but either way you should be able to figure it out.

 

[Glycogen]

I don't have destroyer but the formula you would use to calculate this is deltaE = Break - Make.
So, you calculate the energy of all of the reactant bonds and subtract the energy of all of the bonds in the products made.
Which will give you the amount of energy the reaction releases.

I'm not sure if it is deltaE or deltaH but either way you should be able to figure it out.

I understand that but don't we multiply bond break and bond form by their numbers.For example if we have 6 C-C bonds and 4 C-H bonds don't we multiply bond energies by these numbers!?