bomb calorimeter vs coffee cup calorimeter and enthalpy

[GomerPyle]

Please help me understand the differences and the equations between the two.

A coffee cup calorimeter occurs at constant pressure because it is not completely sealed from atmospheric pressure, and so the pressure will just be the pressure of the atmosphere (volume is free to change in this calorimeter). deltaH = deltaU. U is internal energy.

In a bomb calorimeter, the fluid in reaction is placed in a metal container that is inserted into insulated water. The reaction occurs at constant volume (pressure is allowed to change due to the reaction because it is completely sealed but volume remains constant). deltaH = q, where q is heat.

I want to know why delta H = deltaU in coffee cup, and why it equals heat in bomb calorimeter. The initial equation was deltaH = U + PV. Why does PV cancel in coffee? Where does q come from for bomb?

Thanks guys.

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[Gauss44]

For the differences mathdude2012 on youtube does a great job of explaining.

 

[jtran7]

Please help me understand the differences and the equations between the two.

A coffee cup calorimeter occurs at constant pressure because it is not completely sealed from atmospheric pressure, and so the pressure will just be the pressure of the atmosphere (volume is free to change in this calorimeter). deltaH = deltaU. U is internal energy.

In a bomb calorimeter, the fluid in reaction is placed in a metal container that is inserted into insulated water. The reaction occurs at constant volume (pressure is allowed to change due to the reaction because it is completely sealed but volume remains constant). deltaH = q, where q is heat.

I want to know why delta H = deltaU in coffee cup, and why it equals heat in bomb calorimeter. The initial equation was deltaH = U + PV. Why does PV cancel in coffee? Where does q come from for bomb?

Thanks guys.

internal energy = q + W

q (heat) is basically enthaply

w (work) = Pressure x Volume

the reason why PV cancels out is because there is no volume change resulting in no work (constant P x zero) therefore the change in internal energy is only equal to the heat that is added or removed from the system.

so lets use the bomb calorimeter if you have a chemical reaction that is exothermic (-q) the change in the temp of the water tells you how much q was released via q=m*c*deltaT since the system is closed there is no volume change so there is no work done. As a result the internal energy of the system is less because change in internal energy is equal to negative q + zero work (deltaU = deltaH + W)

as for the coffee cup calorimeter it is a tool that is trying to mimic the bomb calorimeter (a more precise tool) so you would also assume volume does not change resulting in W = 0 and change in internal energy = change in enthaply (heat) only

 

[mehc012]

Please help me understand the differences and the equations between the two.

A coffee cup calorimeter occurs at constant pressure because it is not completely sealed from atmospheric pressure, and so the pressure will just be the pressure of the atmosphere (volume is free to change in this calorimeter). deltaH = deltaU. U is internal energy.

In a bomb calorimeter, the fluid in reaction is placed in a metal container that is inserted into insulated water. The reaction occurs at constant volume (pressure is allowed to change due to the reaction because it is completely sealed but volume remains constant). deltaH = q, where q is heat.

I want to know why delta H = deltaU in coffee cup, and why it equals heat in bomb calorimeter. The initial equation was deltaH = U + PV. Why does PV cancel in coffee? Where does q come from for bomb?

Thanks guys.

This is from another thread where the OP had a similar question. Sorry for not making a new one specifically for your question, but it was a LOT of typing the first time around and I didn't want to repeat it!

ΔH=q only when pressure is constant.

So, the confusing part here is that the first equation is really ΔH = ΔU + Δ(PV)
While this may look the same, it's actually different in that changes in pressure also affect enthalpy. Under constant pressure, it turns out to be the equation you're used to (the one chemists use most often because reactions done on the bench are isobaric):

ΔH = ΔU + Δ(PV)_____Δ(PV) = PΔV if P is constant
ΔH = ΔU + PΔV______ PΔV = W in isobaric conditions
ΔH = ΔU + W
ΔH = Q

However, starting from the full equation yields different results under, say, isochoric conditions:

ΔH =ΔU + Δ(PV)_____Δ(PV) = VΔP if V is constant
ΔH = ΔU + VΔP______ If volume is constant, no work is done, W = 0
ΔH = Q + Δ(PV)_____ Remember that if Q = ΔU + W and W=0, Q = ΔU

The inconsistency is that, in your coffee cup, you're essentially assuming that the volume also does not change (W = 0) because you cannot measure it, and therefore ΔH = ΔU.
In the bomb, W is still definitely 0, so Q =ΔU...but you CAN measure the change in pressure, so you take VΔP into account as well.

 

[GomerPyle]

internal energy = q + W

q (heat) is basically enthaply

w (work) = Pressure x Volume

the reason why PV cancels out is because there is no volume change resulting in no work (constant P x zero) therefore the change in internal energy is only equal to the heat that is added or removed from the system.

so lets use the bomb calorimeter if you have a chemical reaction that is exothermic (-q) the change in the temp of the water tells you how much q was released via q=m*c*deltaT since the system is closed there is no volume change so there is no work done. As a result the internal energy of the system is less because change in internal energy is equal to negative q + zero work (deltaU = deltaH + W)

as for the coffee cup calorimeter it is a tool that is trying to mimic the bomb calorimeter (a more precise tool) so you would also assume volume does not change resulting in W = 0 and change in internal energy = change in enthaply (heat) only

Thanks for the clarification. So the bomb calorimeter is always enthalpy = heat since there is no change in volume (but as the poster said above, there can be change in pressure so how does that take into account?).

For the coffee, the pressure is constant but volume is free to change, but we assume it doesn't so both P*deltaV and deltaP*V are zero?...resulting in work equal to zero and enthalpy (heat) = change in internal energy.

 

[mehc012]

Thanks for the clarification. So the bomb calorimeter is always enthalpy = heat since there is no change in volume (but as the poster said above, there can be change in pressure so how does that take into account?).

For the coffee, the pressure is constant but volume is free to change, but we assume it doesn't so both P*deltaV and deltaP*V are zero?...resulting in work equal to zero and enthalpy (heat) = change in internal energy.

It doesn't matter that pressure changes, because any gas which undergoes 0 volume change does no work. Look at the equations I posted...enthalpy does NOT equal Q in all conditions. It equals the internal energy plus Δ(PV).

ΔH = ΔU + Δ(PV)

It just so happens that, when pressure is constant,

Δ(PV) = PΔV = W

and since, by definition

Q = ΔU + W

we can substitute these in to show that, under isobaric conditions (like the coffee cup)
ΔH = Q


OK, cool. So, to repeat, that is COFFEE CUP, or ISOBARIC.

Now we're going to look at BOMB CALORIMETRY, or ISOCHORIC:

Remember, the initial equation is

ΔH = ΔU + Δ(PV)

This time, volume is constant, and so

Δ(PV) = VΔP

However, by definition, as the gas is neither expanding nor contracting, W = 0. Think back to our basic work/energy/heat equation for a gas:

Q = ΔU + W

If work is zero, then in this case, Q = ΔU, and the two can be used interchangeably.
When we put these equations together, we are left with

ΔH = Q + VΔP or, if you prefer
ΔH = ΔU + VΔP
the two are equivalent under isochoric conditions.

 

[GomerPyle]

It doesn't matter that pressure changes, because any gas which undergoes 0 volume change does no work. Look at the equations I posted...enthalpy does NOT equal Q in all conditions. It equals the internal energy plus Δ(PV).

ΔH = ΔU + Δ(PV)

It just so happens that, when pressure is constant,

Δ(PV) = PΔV = W

and since, by definition

Q = ΔU + W

we can substitute these in to show that, under isobaric conditions (like the coffee cup)
ΔH = Q


OK, cool. So, to repeat, that is COFFEE CUP, or ISOBARIC.

Now we're going to look at BOMB CALORIMETRY, or ISOCHORIC:

Remember, the initial equation is

ΔH = ΔU + Δ(PV)

This time, volume is constant, and so

Δ(PV) = VΔP

However, by definition, as the gas is neither expanding nor contracting, W = 0. Think back to our basic work/energy/heat equation for a gas:

Q = ΔU + W

If work is zero, then in this case, Q = ΔU, and the two can be used interchangeably.
When we put these equations together, we are left with

ΔH = Q + VΔP or, if you prefer
ΔH = ΔU + VΔP
the two are equivalent under isochoric conditions.

Awesome - thank you very much.