Okay so I believe this is a fairly straight forward enthalpy problem, but for what ever reason it says i'm wrong.

 

You carry out an exothermic reaction inside a Styrofoam cup calorimeter, using a temperature probe to monitor the temperature throughout the reaction. You find that the temperature increases 6.37 °C during the reaction. The specific heat of the reaction solution is known to be 6.20 J/g•°C, and the mass of the solution was measured as 33.0 g. Calculate q, the heat for this reaction, in kJ. Be sure to include the correct sign for your answer.

 

So here is what I did.

I understand the equation i'm to use is q=cm(change in temperature)= (specific heat of solution)(mass of solution)(change in temperature).

So by following this formula I got;

 

q=(33.0g)(6.20 J/ g centigrade)(6.37 centigrade)(1 kJ/ 1000 J) = 1.30 kJ

 

Did I miss something completely or is there an error in my conversions?

Hello Ember - this is not really my scene but a couple of things immediately sprang to mind on these sorts of questions.

 

1. Accuracy - are you putting too many decimal places?

2. The line in the question "Be sure to include the correct sign for your answer." is a dead giveaway that they might be setting you up for a fall!

 

Take a read of this link on Exothermic and Endothermic reactions

Okay so I believe this is a fairly straight forward enthalpy problem, but for what ever reason it says i'm wrong.

 

Did I miss something completely or is there an error in my conversions?

Before doing this experiment, have you done any experiment about the reference reagents by using the same apparatus?

Quantitative experiment always should be exact, even the experiment is in the elementary course.

alpha2cen, this is probably a question from a textbook. I don't think that there is any chance to to additional tests.

 

on topic:

I think that the convention is that heat that you put into a system gets a positive sign (it is the heat required), and heat produced gets a negative (you have a negative heat requirement). Since your sample heats up, the Q should probably get a negative sign. But this always confuses the hell out of me, so in my work, I prefer just to avoid the symbol Q altogether, and just explain in normal words what happens: normally stuff either heats up or cools down, and needs a certain amount of fuel, electricity or whatever to do that...