This is on a practise exam I have and I cant get part of it, I don't know if it is not only due to my not understanding but to the fact that the first two parts might be wrong. Thanks for your help if you can. Heres the question:
A coffee-cup calorimeter normally consists of two nested styrofoam cups with a lid. A coffee-cup calorimeter of this type contains 125.0 g of water at 20.00 degrees C. A 120.0 g sample of copper metal is heated to 98.50 degrees C by putting it into a beaker of hot water. The copper is then put into the calorimeter and the final temperature of the water and copper is 25.80 degrees C.
A) Calculate the energy change of the copper metal.
For this answer I got -3358.74J
B) Calculate the energy change of the water.
For this I got 3033.40J
C) The difference between the two values is due to energy lost to the styrofoam cups. The heat capacity of a calorimeter is the amount of energy required to change the temperature of the apparatus by 1 degree C. Calculate the heat capacity of the calorimeter in J/degrees C.
This one im stuck.
D) In a typical coffee-cup calorimeter we assumer that the energy change due to the calorimeter is negligible. In this experiment is this a vaild assumption? Exaplin.
I;m assuming this might be easier to asnswer once I get part C
Your answer to A is ok as is B.
For C, heat capacity = q/delta T.
I will let you ponder D.
Oh I know that much, sorry I didn't clarify that, what I don't know is, do you subtract A from B to get q or vice versa?
Would it matter? You know heat is being added to the cup so you know heat difference must be + so subtract "any old way" and stick a + sign in front of the difference. However, if you want an equation remember that
EH2O + ECu + Ecup = 0
3033.4J - 3358.74 + heatcap x delta T = 0
To clarify with parenthese:
EH2O + ECu + Ecup = 0
3033.4J - 3358.74 + (heatcap x delta T) = 0
To calculate the heat capacity of the calorimeter (part C), you can use the equation:
EH2O + ECu + Ecup = 0
Since the energy change of water (EH2O) is 3033.4 J (as you calculated in part B) and the energy change of copper (ECu) is -3358.74 J (as you calculated in part A), you can substitute these values into the equation:
3033.4 J - 3358.74 J + (heatcap x delta T) = 0
Now, we can rearrange the equation to solve for the heat capacity (heatcap):
heatcap x delta T = 3358.74 J - 3033.4 J
heatcap x delta T = 325.34 J
Since the delta T (change in temperature) is the final temperature minus the initial temperature, we have:
heatcap x (25.80 degrees C - 20.00 degrees C) = 325.34 J
heatcap x 5.8 degrees C = 325.34 J
Finally, we can solve for the heat capacity:
heatcap = 325.34 J / 5.8 degrees C
heatcap ≈ 56.13 J/degrees C
So, the heat capacity of the calorimeter is approximately 56.13 J/degrees C.
Now, moving on to part D, in a typical coffee-cup calorimeter, we assume that the energy change due to the calorimeter is negligible. This assumption is based on the idea that the heat capacity of the calorimeter is small compared to the heat capacity of the substances being measured (water, copper, etc.). In this experiment, the calculated heat capacity of the calorimeter (56.13 J/degrees C) is relatively small compared to the energy changes of water (3033.4 J) and copper (-3358.74 J), so the assumption is valid.