# Calculating Heat Capacity of Calorimeter

• Armitage12

#### Armitage12

Need a little help, as I seem to have gotten confused.
Looking over past exam questions for the heat capacity of a calorimeter, this one is the one I am looking at :

A sample of the sugar fructose (C6H12O6) of mass 0.900 gwas placed in a calorimeter and
ignited inthe presence of excess oxygen, at the saturation vapour pressure of water and at
constant volume.The temperature rose by 1.4 K.
The calorimeter had previously been calibrated by passing an electrical current through a heating
tape for 100 s. The power supplied to the tape was 500 W and the temperature of the calorimeter
rose by 5 K

Right. The reason I am confused, is the textbook I've been reading from had similar worked examples, however they didn't have any values in watts. They have in Ampere and Volts.
As far as I am aware 1 watt is equal to 1j/s..

So to work the heat capacity of the calorimeter,
would i:

1) 500W*60 to get to 8.3 J/s
2) then multiply by 100s and divide by the 5K.
3) then multiply this answer by 1.4k, to get 233.3 joules.
4) Finally, divide by 1000, to get the heat capacity of the calorimeter at 0.23KJ?

Or am I way off?

Or am I way off?
Waayyy ooblackf.
heating tape for 100 s
This is a clue.
500 W
This is another clue. Hint, what's "Watt" mean, what are the units?
rose by 5 K
Then, divide.

The units of W is one Joule per second?, so If i multiply the 500W by the 100s. I get 50000 J?
Do I then divide this by the 5K? Finally multiply by the 1.4k? to get the final answer at 14Kj?

Waayyy ooblackf.

This is a clue.

This is another clue. Hint, what's "Watt" mean, what are the units?

Then, divide.

The units of W is one Joule per second?, so If i multiply the 500W by the 100s. I get 50000 J?
Do I then divide this by the 5K? Finally multiply by the 1.4k? to get the final answer at 14Kj?

Or do I need not multiply by the 1.4k.

Is it simply 500w x 100s/ 5k= 10,000J/1000= 10kJ??

Why so uncertain?
Upper case "K" denotes Kelvin, or degrees Kelvin (same size as C); you'll want lower case "k" to denote "kilo" or thousand; lower case "j" has a number of meanings, but none indicate Joule, or upper case "J." Other than that, you're in business.

Or do I need not multiply by the 1.4k.
Now I know why you're uncertain. What's the definition of "heat capacity?"

Now I know why you're uncertain. What's the definition of "heat capacity?"

The amount of heat needed to raise the temperature of an object or substance by 1 degree.. So I don't need to multiply by 1.4k at the end?
Just 500W x 100s/5k= Heat capacity of Calorimeter? so 10 kJ?

Sorry its been a long day haha.

heat needed to raise the temperature of an object or substance by 1 degree
Think. What are the units? You multiplied watts by seconds and got joules. If you divide joules by temperature, what do you get for units?

I would of assumed J/K?

Yes. Now, you know the heat capacity of the calorimeter. You know the temperature rise for a particular measurement. How do you calculate the heat released in that measurement?

Yes. Now, you know the heat capacity of the calorimeter. You know the temperature rise for a particular measurement. How do you calculate the heat released in that measurement?

So now I just multiply it by the 1.4K? to get the amount of heat released as I had it in J/K. So as it was in Joules per kelvin, if i times by 1.4 Kelvin, the heat will adjust, so I will know how much?

But the question was only asking for the heat capacity of the calorimeter? not how much heat was released? So I am right in assuming, that for just the heat capacity of the calorimeter, I don't need to multiply by the 1.4K?

Correct.

Correct.
You were a massive help, really appreciate it!