Constant-Pressure Calorimetry - Video Tutorials & Practice Problems
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Constant-Pressure Calorimetry uses a coffee cup calorimeter to determine the heat transfers occurring in a liquid solution.
Constant Pressure Calorimetry
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concept
Constant-Pressure Calorimetry
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Now a calorimeter is just a container usually insulated against heat loss that contains a liquid with a given heat capacity. We're going to say usually when a heated object is placed in a liquid, both the liquid and the container that it's in will absorb this released heat. And we're gonna say here that our standard heat capacity which is capital c for a calorimeter, it's just the amount of heat required to change its temperature. Now the weight of the calorimeter is usually unknown so we're gonna exclude it from our our mass in terms of the formula. Now standard heat capacity equals capital c equals q, which is our heat, divided by change in temperature. So our standard heat capacity is usually in units of joules over degrees Celsius. Now this temperature could be given to us in Kelvin and if it is given to us in Kelvin, then the temperature must be converted to Kelvin as well. The units here both have to match one another. Here, our q which is our heat can be in joules or in kilojoules. So always be on the lookout for that too. Make sure your units all match so that they cancel out to isolate the missing variable.
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example
Constant-Pressure Calorimetry Example 1
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What is the standard heat capacity in kilojoules per degree Celsius of a calorimeter that absorbs 87 joules at its temperature as its temperature goes from 33 degrees Celsius to 38.1 degrees Celsius. Alright. So your standard heat capacity is c, capital c, and remember it equals q divided by change in temperature. Here we need to convert our joules into kilojoules because they want the unit in kilojoules. So we have 87.0 joules. Remember that 1 kilo is 10 to the 3 joules here. So that's 0.087 kilojoules. Take that and plug it in. And then delta t is our change in temperature. Final temperature minus initial temperature. So when we do that, we're gonna get our answer in kilojoules over degrees Celsius. This comes out to be 0.01 71 kilojoules over degrees Celsius. Here, our answer is gonna be given to us in 3 sig figs because I'm using 87.0 joules as my reference. Here our temperatures could also be used. This has 2 significant figures. This has 3. But just to give more detail into our answer, we're gonna go with 3 significant figures.
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concept
Constant-Pressure Calorimetry
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Now constant pressure calorimetry is associated with our coffee cup calorimeter. Now we're going to say here, it uses the coffee cup calorimeter to determine heat transfers occurring in a liquid solution. Now a coffee cup calorimeter is basically an insulated styrofoam cup with a lid, and we're gonna say we call it constant pressure because the calorimeter measuring heat is open to the atmosphere where pressure is going to be fixed. So here, if we're gonna take a look at the images provided to us, we're gonna start out with our coffee cup calorimeter here on the left. And when we look at this coffee cup calorimeter, we're gonna say it has some key portions to it, one of them being this thermometer given to us on top. So that's gonna measure my temperature difference. Then we're gonna have here our styrofoam cover or styrofoam top. Here, this portion just represents the styrofoam cup itself. In here we have our water, and then finally here we have our stirrer. When we place the heated object inside of here, we're gonna be able to measure the amount of heat that's being released. Now realize here when it comes to the cup, constant pressure calorimeter and this coffee cup calorimeter, we have formulas that are associated with it. We're gonna say when both the liquid and calorimeter absorb heat from the hot object we get the heat lost by the object, so minus q, equals plus heat gained by the water, plus the heat also gained by the calorimeter. Now we can expand this a little bit further. We're gonna say expanding them to their heat capacity formulas where q equals mcat. We're gonna have q lost is minus mcat, q gained by the water is positive mcat. And remember, for the calorimeter, the mass of it is usually unknown so we ignore its mass within our formula. So it just becomes the heat capacity of the calorimeter which is capital c times change in temperature. So that becomes our equation for constant pressure calorimetry. It's going to be minus mcat equals plus mcat plus the specific heat well, the heat capacity of the calorimeter times its change in temperature.
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example
Constant-Pressure Calorimetry Example 2
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Here we're told when 60 grams of lead at 68.3 degrees Celsius is poured into 90 grams water at 30 degrees Celsius within a coffee cup calorimeter, the temperature increases to 48 degrees Celsius. Based on this information, what is the heat capacity of the calorimeter? We're told the specific heat specific heat of lead and water are 0.128 and 4.184 respectively. Alright. So we're gonna plug well, first of all, we need to realize we have our container with water and we're placing it placing into it our lead. Lead is at a higher temperature than the water is. Remember, the hotter object always releases its heat. So it's gonna release its heat. So the water is releasing its heat, so negative q of water or no negative q of lead, sorry. The object that's releasing its heat equals positive q of water which is absorbing it, plus q of the calorimeter. The calorimeter is absorbing some of it as well because they tell us that we need to find a t capacity. We don't need to find a t capacity if it was involved within the calculation. Alright. So if their q's are involved that means their m cats for some of them are involved. So negative m cat o r lead equals positive m cat of the water. Remember, calorimeters, their masses tend to be unknown so it's just the heat capacity that they have times the change in temperature. Alright. So here we're gonna say negative 60 grams of our lead times its specific heat capacity times the change in temperature. We're told that the temperature increases to 48 degrees Celsius, that's the final temperature for everyone. So here, the change in temperature for the lead is the final temperature minus its initial temperature of 68 0.3 degrees Celsius. That equals positive mcat of water, mcat on so water is 90 grams of water, times its specific heat capacity times its final temperature minus the initial temperature that it had, plus we don't know the heat capacity of the calorimeter, that's what we're looking for. And here's the thing, before we add the lead to the water, it was just the water in the calorimeter hanging out with each other. So they both must have been at the same initial temperature. So change in temperature would be 48 degrees Celsius minus the initial temperature of the calorimeter, which again, it was with the water and we assume they both were at the same initial temperature of 30 degrees Celsius. All we have to do now is solve for this one missing variable. So what we're gonna do now is we're going to multiply everything here on this side, Grams will cancel out, degrees Celsius will cancel out. We'll have joules left on this side which comes out to 155.904 joules equals grams cancel out, degrees Celsius cancel out. So we're gonna have everything here also in joules. So that is 6 778.08 joules plus, when you subtract these, that's gonna give me 18 degrees Celsius times my specific heat capacity or my heat capacity of c. Got us we gotta isolate our heat capacity of c. So next we're gonna subtract this from both sides here. So when we do that we're gonna have now negative 66.22176 oh, actually negative 6622.176 equals 18 degrees Celsius times my heat capacity of c, divide out 18.0 degrees Celsius, so remember joules are still here, so 18.0 degrees Celsius. Heat capacity is a number that is positive, so this is in absolute terms. When we do that, that's joules divided by degrees Celsius. So here, my heat capacity, which is capital c equals 367.9 joules over degrees Celsius. So that would be the heat capacity of our calorimeter within this given question.
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Problem
Problem
A 115.6 g piece of copper metal at 182.5 ºC is placed into 120.0 mL of methylene chloride at 31.0 ºC within a coffee-cup calorimeter. If the final temperature of the solution is 50.3 ºC, what is the specific heat of methylene chloride? Assume the calorimeter absorbs a negligible amount of heat. The specific heat of copper is 4.184 J/g ∙ ºC and the density of methylene chloride is 1.33 g/cm3.
A
5.022 J/g°C
B
8.178 J/g°C
C
12.94 J/g°C
D
17.28 J/g°C
E
20.76 J/g°C
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Problem
Problem
You place 75.0 mL of 0.100 M NaOH in a calorimeter at 25.00 ºC and carefully add 55.0 mL of 0.200 M HNO3, also at 25.00 ºC. After stirring, the final temperature is 53.35 ºC. Calculate the enthalpy ∆Hrxn (in J/mol) for the formation of water. (Specific heat capacity, Cs, and density of the solution:4.184 J/g∙K and 1.00 g/mL).
A
-9.25x10-3 J/mol
B
-7.72x10-5 J/mol
C
-4.57x10-5 J/mol
D
-2.05x106 J/mol
E
-6.39x10-8 J/mol
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