Hi!

I completely understand where you are at as far as being scared about Chem. 101 and wanting so bad to get into a nursing program! I am a couple days away from taking the Final in my Chem. 101 class and I am sooo scared!

But I don't mean to scare you any further. First, you should know that with the help of a tutor (do they have tutors at your school?) or a classmate whose got the hang of it (I just know there is at least one) you can totally learn to solve these questions, so don't panic just yet! I definetely understand that thats easier said than done (I developed a stress-based stomach ache over a Chem. quiz this morning lol - thank goodness the teacher called it off!). But really don't worry so much - but DO get help all quarter long! Trust me when I tell you its the key to success in Chem. 101 - that and studying all the time. With all that said I will now show you how to solve your bottom-most problem:
First of all, as you might already know, problems like these use the following equation:
q = m x C x △T
They also use a type of measurement called Joules (read all about it in your textbook)
q means the the end result of this equation and it is the amount of heat in Joules (I will have figure out what this is going to be in this problem)
m is the mass of the substance in grams (In this problem that is 168 g) So we already know that the mass is 168g
C is the specific heat (this is what we are trying to find)
△T is the temperature change in celsius, and to find this we subtract the high temp. from the low temp. (so in this problem we do this: 74 - 26 = 48) Now we know that the temp. change is 48 celsius
You have probably already noticed that we need to solve for the center part of the problem, that is C (specific heat), but you are might be wondering what q is. Here is how we solve for q:
we have 2750 calories of heat. It just so happens that 1 cal. of heat = 4.18 Joules
Recall that q is Joules of heat. So set up the problem using the cal to Joules ratio as a conversion factor. 2750 cal. times 4.18 J. divided by 1 cal. : 2750 cal (4.18 / 1)
From this we find that 2750 cal. of heat = 11495 Joules of heat! Now we have solved for q and it is: q = 11495 Joules
So now we need to solve for C (specific heat) - the whole big question of our problem.
Equation is now: 11495 Joules = 168 g. x C x 48
Now using the logic of simple algebra, we realize that to figure out C we have to isolate it on one side. We do that by dividing both sides by 168 x 48. This isolates C.
Equation is now: 11495 / (168 x 48) = C --- turns into: 11495 / 8064 = C -----
turns into: 1.42547123 = C --- now we have solved for C! Round the answer to the correct amount of spaces and our answer is:
1.43 C (specific heat)
This should be the correct answer, because when I plugged it back into the equation it worked.
I sure hope this explanation helped (sorry its so long winded) and I'll look into how to solve your other question (that looks a little tougher). You should learn this type of equation through and through because the teacher will ask you to solve for different parts of it in different problems (sometimes for q, sometimes for m, sometimes for C, and sometimes for △T). I wish you ALL the best of luck! Let me know if this helped and if you have anymore questions. Don't worry, you can get a good grade in Chem. 101, you just have to get some help like the previous poster said.
Originally Posted by twinjeep97
Hello, I am taking Fundamentals of chemistry this summer and I started today. I am already having problems with this class! The teacher goes speedy fast, and has no time to explain anything. I need to do well in order to apply for the nursing program. Please , I am begging and pleeding anyone and everyone to help.
I am having problems with the math part of chemistry. Example:
"A European car advertises an efficiency of 22 km/L, while an American car claims an economy of 30 miles/gal. What car is more efficient?"
OR
"If 168g of an unknown liquid requires 2750 cal of heat to raise it's temperature from 26 degrees celsius to 74 degrees celsius, what is the specific heat of the liquid?"
I am lost as to were to start on these problems. I need help,please anyone! Thanks so much!