.
Hello, Abigail (and Stanton D.),
Stanton is correct and he has given you the best advice on how to proceed. When he says "follow the units," just think about what is being asked and what data you have. In your problem it is a little more difficult until you think through what is happening. For me, it helps to rewrite, in a table, what I know to get started. But most important is to understand what is being asked.
The question is what will the temperature change of the calorimeter be if we combust a "certain" sample of cyclohexane in which 9892 J of energy is released. [I never know why "certain" is used to describe samples]. So we know that the problem is one in which we need to understand the definitions of heat capacity and specific heat.
Let's start with specific heat, which is defined as the amount of heat required to raise the temperature of an amount of substance by 1 degree. A large specific heat means it takes more heat to raise it by one degree.
We can represent this value in many different units, which is often a key hurdle in understanding and completing a problem. That's one reason why Stanton advises to focus on the units. I agree.
Look at two examples of specific heats of different materials to better understand what specific heat tells us;
Iron : 0.444 J/goC
Water: 4.184 J/goC
Most people, when asked "Which can carry the most heat?" will respond "iron." But these numbers tell us that it is water, by a long shot. For 1 gram of each, it takes almost ten times the amount of heat to raise water by 1 degree C than iron. It makes sense when you realize what this means. Iron does not tolerate heat very well. It's temperature rises quickly when heat is applied. Water, on the other hand, is a voracious consumer of energy. "Bring it on" is it's response top heat, and it rises a degree only after a lot more energy is applied. So an iron skillet is good for cooking (the heat goes right through) and it takes forever and a day waiting for water to boil.
Heat capacity is a measure of how much heat required for an object to increase by one degree. That's what is given to us for the heat calorimeter. It is the heat needed to raise the whole calorimeter by one degree. So it doesn't have a mass unit in the number. That makes the problem a lot easier than it would first appear, and the reason for taking time to understand the question. The next section, however, explains more of the concepts before I get back to this specific question.
Notice the units of J/goC. All three can measures can be converted into different units, so we can have specific heats such as J/moleoC, J/goK, kJ/goC, cal/goC, and so forth. But despite the unit differences, all are valid expressions. One simply has to pick the one most useful for the data or convert the units as necessary.
Specific heat can be written as an equation: c = q/mDt, where
c = specific heat
q = heat/energy
m = mass (or equivalent measure such as moles)
and Dt (Delta T) which is the temperature change (T1 - T2). T1 is initial and T2 is final.
When specific heats are reported, they are always in 1 degree (C, F or K) amounts. In many problems, we want to know the temperature change, so Dt would be in the expression.
I usually make a list of the data, along with the units. But the most important step is to understand the question. Heat is being released into a bomb calorimeter that can absorb that heat in a manner that for every kiloJoule (kJ) it's temperature will rise by 1 oC. It is expressed as kJ/oC. In this case there is no mass used to describe the heat capacity (it is not the specific heat) of the calorimeter, so mass is not in the unit.
Stanton D. read the question first, so he saw the ease in which this problem could be solved. He suggests looking at the units, so let's do that. We have 9892 Joules of energy being absorbed by a calorimeter that can take the heat by raising its temperature by 1 oC for every 7.4 kJ. If we want degrees C as the final answer, then think about how we can use these two data points to produce a value of degrees C. I can see that by dividing one by the other, I can cancel energy units, if only they were the same. So convert one into the other (e.g., change kJ to J by multiplying by 1,000; or change J to kJ by dividing by 1,000).
Then think about the units at the same time making some guess as to what should happen (go up, down, negative, etc) so that the final result can be checked against your expectations.
I hope this helps explain Stanton's wise suggestion to follow the units. But also keep in mind what the experiment is trying to tell us.
Bob
Abigail E.
I am confused about how to go through the process, can you help me work through it?10/07/20