When a piece of Copper weighing 250 grams is placed in a cup with 450 ml of water at 21OC and the Cp of the cup is 47 J/K, how many grams of gasoline would it take to heat the system to 110OC?
See your previous post under Cynthia.
It's Jessica from my first screen name
On this problem I wanted to know if I'm going with the correct formula
q= CpxmassxTf-Ti
250 g being mass I just do t know how to set up the formula
The part you have is ok but you've posted only part of what you need.
First, I assume that 21O C and 110O C are 21o C and 110o C.
Now you have Cu wire + water as a system.
You need to calculate q for the copper wire and add to q from the water and add to q for the cup and add to q for water turning to steam and add to q for raising steam T from 100 to 110. That will give you q needed from the gasoline. Then you look up the heat of combustion for gasoline and go from there.
For example, the equation you wrote will take care of the Cu part; i.e.,
qCu = mass Cu x specific heat Cu x (Tf-Ti) = 250 x Cp Cu x (110-21) = ?
qCup = 47*(110-21) = ?
q for heating H2O from 21 to 100 is
q = mass H2O x specific heat H2O x (110-21) = ?
q for turning H2O to steam is
a = mass H2O x heat vaporization = ?
q for raising T H2O from 100 to 110 is
q = mass H2O x speicif heat steam x (110-100) = ?
etc.
Cu specific heat: 0.385
H2o specific heat: 4.184
21+273.15=294.15
110+273.15=383.15
{I subtracted these and got}
______
89
250x0.385(383.15-294.15)= 8566.25
450x4.184(383.15-294.15)= 167569.2
47(110-21)=4183
I'm lost on this problem
As I said before, this is a multi-step problem. What you're doing is calculating how much heat is needed for each of the steps in getting Cu from 21 to 110 and in getting the water and the cup from 21 to 110. The next part of the problem then is to calculate the amount of gasoline needed to supply that amount of heat to the Cu, cup, and water.
Cu specific heat: 0.385
H2o specific heat: 4.184
21+273.15=294.15
110+273.15=383.15
{I subtracted these and got}
______
89 And you can get 89 also by 110-21 = 89 so in these problems it isn't necessary to convert to K and subtract; it's much easier to keep it in C and subtract. You can that legally because 1 degree C and 1 degree K are the same.
250x0.385(383.15-294.15)= 8566.25 This is the heat needed to raise the T of the Cu from 21 to 110.
450x4.184(383.15-294.15)= 167569.2
This is the heat needed to raise the T of the 450 mL H2O from 21 to 100.
47(110-21)=4183 This is the heat needed to raise the T of the cup from 21 to 100.
Next you calculate the heat needed to convert the 450 mL liquid water to steam at 100 C. I gave you that formula earlier.
Also, add heat needed to raise the T of the steam from 100 to 110 C. I gave you that formula earlier.
All of them added together is the heat needed to be supplied by the combustion of gasoline.
I made a typo on your "cup" line. Here is the correct copy. The 100 C should be 110 C.
47(110-21)=4183 This is the heat needed to raise the T of the cup from 21 to 100.
450x4.184(100)188280
450x4.184(110-100)= 18828
18828+4183=23011
23011g of gasoline
450x4.184(100)188280 I don't know what this is. It isn't anywhere in the above. It should be 450 x 4.184 x (100-21) = ? if it's the heat needed to raise the temperature of the water from 21 to 100.
450x4.184(110-100)= 18828 I assume this is supposed to be the heat needed to raise the temperature of the steam from 100 to 110; however, 4.184 is NOT the specific heat of steam.
18828+4183=23011 THis is the sum but it doesn't contain all of the different q values you should have calculated. There is no value for Cu being heated, no value for the cup being heated, no value for the vaporization of water. You should have five of those q values. Add all of them together.
Finally, you have
23011g of gasoline which I assume means you think that joules of heat needed to get all of those values is the grams gasoline needed but that isn't so. After you arrive at the total q value, look up the value for the combustion of gasoline. It will be quoted as so many kJ/mol. We can talk about how to do that part when you get to that part.
Steam :
450x1.86(110-100)=8370
8566.25+18828+8370+4183+89=40036.25kj
(40036.25)(114.2g/mol)=
4572139