5.2 Calorimetry Flashcards
Calorimetry is the studies of heats of rxns, or enthalpys of rxns (change in systems heat)
Calorimeter - insolated device that is designed to study the heat of rxns
* Amount of heat produced by a rxn
* its a very heavily insulated container (minimizes exhgange of heat w/ surroundings)
* You would stick some sample in the calorimeter, and do some rxn on the inside and measure the heat change inside the calorimeter to see how much heat was produced during the rxn
* This is actually how they estimate how many calories food has
* They will light the food on fire and let it burn completely and see how much the temperature changes
* So why can they light it on fire in the calorimeter and have it burn there and be the same as your body consuming it? Well enthalpy change is a state function - only dpeends on the inital and final state, it doesnt matter what path it takes to get there. its actually basically a combusion rxn within your body, which is the same net result as lighting it on fire
Calorimeter
* q = C(sub cal) * Delta T
* heat capcity = C (how much heat it takes to change the temperature of the calorimeter) - The amount of heat required to raise the temperature of the caloremeter 1C - or tells you how much E is needed to raise the entires samples temp by 1C
* q = heat
* Delta T = the change in temperature
Typically you figure out what the heat capcity is w/ a controlled experiment and you calibrate it in that way.
If 3000J of heat causes a temperature increase of 3.0 C for a calorimeter, what is its heat capcity?
If 8g of methan (CH4) is combusted in a calormeter w/ a heat capcity of 29.7 kj/C and its temp rises from 25.0 C to 40.0 C, what is the enthalpy of conbustion of methane in kj/mol
1 calorie = 1 calorie to raise 1g of liquid water 1 degree
Specific heat = the amount of heat it takes to raise 1 g of a substance 1 degree C
Specific heat equation:
* q = m * Cs * Delta T
* q = amount of heat
* m = mass
* Cs = specific heat of substance
* Delta T = change in temperature
The idea w/ have mass times specific heat cacpcity is specific heat capcity is specific to 1g of the substance, so we have to multiply it by the entire weight of the object to get the full specific heat
w/ a specific heat you have a specific substance your using, and when you add heat to it, the temperature goes up, and when you remove heat from it its temperature goes down, and thats the realtionship were looking at here. So if you’ve got a certain amount of a specific substance you’re doing a specific heat problem, not a calorimeter problem.
Think q = MCAT - kind of looks the same as the equation above - good way to remember it
You need to be specific about what phase you’re in, because they’re all going to have different specific heats.
* Again the definition of specific heat is how much heat must be added to your specific substance to raise 1g of that substance 1 degree C
* For liquid water that specific heat is 4.18 J / g * C - so if you have 1g of water, its going to take the addition of 4.18J of heat to raise 1g of that water 1 degree C
* If you wanted to raise it 2 degrees than you double it - thats why we multiply by delta T
* if you had 5g of water, than it would take 5x as much heat to raise it 1 degree celcius as 1 gram of water –> this is why we also multiply by the mass in the equation above
With a lower specific heat, it takes less heat to raise its temperature 1 degree celcius
* so think a park bench vs a side walk in the hot summer, the specific heat of that park bench is lower, meaning the temperature of it can be deviated up more easily
* Lower specific heat means it takes less energy to change the temperature - so that objects temperature will fluctuate more easily when gaining or losing heat
* Specific heat = the amount of heat (energy) required to raise the temperature of 1 gram of the substance 1 degree - if it takes less heat to raise 1 gram of something 1 degree that means it has a lower specific heat, meaning it will be more easy to deviate that substances temperature
Heating curve (to the right) is what happens when you add heat to a substance
* So in general we think, when we add heat to a substance the temperature of the substance is going to go up. And usually thats true, except during phase changes
When you add heat to a substance, (q shown below) and its temperature stays constant, thats indicitive a phase change
* The phase changes are shown below in blue
So MCdeltaT allows us to calculate heat changes as long as we don’t pass through 1 of the phase changes
* because in the phase changes the heat is increasing while the temperature remains stable
* so if we pass through one of these phase changes theres no delta T, meaning the equation MCdeltaT will just be 0 and not help us because our temperature is not deviating (even though the amount of heat we put in is obivsouuly changing)
So we would have to look up some thermodynamic values:
* If were pasing through from solid to liquid we would have to look up the delta H of fusion, and if were passing from liquid to gas we would have to look up the delta H of vaporization
Normally that Delta H fusion is provided and kj/mol, so its a scaled value for 1 mole, and you’ll multiply by n, or the number of moles that are in your substance. So if you have 2 moles you take 2 * delta H of fusion
* they may also be reported in units of kj/g, meaning you’d have multiply by number of grams (which might mean you end up converting your moles to grams - ez pz)
So basically everytime you hit one of those phase changes on the heating curve, you have another step in the calcualtion that you’re going to perform
One last note is that the line for the delta H of vaporization should’ve been longer than the line for delta H of fusion. The reason for that is because it takes more heat to boil a substance than it does to melt a substance (takes more heat for that phase change to happen)
What q =MCdeltaT actaully does
This equation calculates the amount of heat energy (q) that is
* absorbed (if temp increases) or
* released (if temp decreases)
How much heat would be required to raise 90g of ice at -20 C to 50C
Well we can see that because this is water, its going to pass a phase change at 0C (going from solid to liquid), meaning we can’t only use q = MC * deltaT to calculate the heat required
So this is going to break down into a 3 part problem
* Going to have to calculate the MCDeltaT from -20 –> 0
* calculate the phase change portion at 0C (delta H of fusion part)
* calculate the MCdeltaT from 0–>50
The solid, liquid, and gas phases of water all have different specific heat values (C), so you’re going to have to plug in those numbers (shown on the next slide)
Thermodynamic data for water (will be plug n chug)
So we must note that were doing a calorimeter problem, not a specific heat problem. So we dont use the MCdeltaT equation
q = C(sub cal) * Delta T
C = Heat capcity - tells you how much energy is need to raise the entire samples temp by 1 degree C
So if the heat capcity of the caloriometer is 210J/C that means that it takes 210J to heat the caloriometer 1C
* So no matter what the caloriometer is made of or how much it wieghts, as a system, it needs 210 J to go up by 1 C
So when solving for heat below, if we want to raise it by 5C it makes since that we would just multiply that 210J/C by C to find q
Difference between specific heat and heat capcity
* Specific heat = Tells you how much E is needed to raise 1g of a substance by 1C - think its specific the the gram amount
* Heat Capacity = Tells you how much energy is needed to raise the entire samples temp by 1C - this is waht we use in caloremeter problems
So this is a specific heat problem so we use q = MCdeltaT
Remember, specific heat is how much E it takes to raise 1g of the substance by 1 degree C. In this problem its asking about 10g, so it makes since that were multiplying by the mass (because if not it would be assuming 1g). and were multiplying by the temperature because specific heat assumes 1 degree C, whereas our sample cahnged by 30 degrees, so its going to take much more E to deviate by 30 degrees than to deviate by 1 degree.
Delta H fusion = enthalpy of fusion - the amount of heat needed to melt 1 mole of a substance at its melting point w/o changing temp
* Enthalpy change = fancy chemistry word for heat at constant pressure
So basically this problem is asking how much energy (q) it takes to melt that ice right at its meltoing point (that long plateue on the graph shown before - where energy is required but temperature doesnt deviate).
Delta H fusion = specific to how much q is required to melt 1 mol of stance substance at its melting point, but the temperature of the entire thing isnt changing (which is why we can’t use q = MCdeltaT
since we had 3 mols we had to multiply the delta fusion by 3 because its specific to 1 mol (which is the equation i utilized below)
So this problem is asking how much heat/energy is required to get ice passed that plateu at freezing. So how much heat is required to deivate it passed that freezing plateu w/o temperature changing
Delta H fusion = 6009J/mol
* that means for every 1 mol of H2O it takes 6009J to get it passed that plateau
* since we have 4 mols, that means that we must multiply this number by 4
* thats why I utilized the equation q = n * deltaH
q = 24000J
This is a good example problem
So in this problem water is the surroundings and iron is the system
This is an exothermic reaction because the extenral environment gained heat, meaning the internal environment (iron) lost porportionaly the same
If water (surroundings) gained 2257.2J of heat, that must mean that iron (system) lost 2257.2J of heat
* Answer = -2257.2J heat
The key here is creating the conversion factor
You can relate 2mol H2 to -484 kj
* meaning for every 2 moles of H2 used -484 kj happens
-484 kj = heat is released
484 kj is released for 2 mols H2
little confused on the signs for this one
When you see a -delta H next to a rxn it means the reaction is exothermic - the system is losing heat and the surroundings are gaining
Delta H is the change in enthalpy (change in heat at constant pressure)
So if you see a reaction with -Delta H = 220 kj, for example, it just means 220 kj of heat is released when that reaction occurs
* which makes since, the system is losing E and the surroundings are gaining it, which is why its written w/ a neagtive infront of it (everything is in the systems prospective)
So the negative delta H means it was an exothermic rxn. Heat was transfered from the system to the surroundings w/ this xn.
* so E was released
So the hard part was seeing that a positive 1310 was released into the environment. It is negative delta H, however, that just means released, so 1310 was released.
* Then for every 1 mol C2H2 1310kj of E was released, which was my conversion factor
