[jillian2577]

I did a lab in chemistry where I had to burn one gram of wood under a metal can and keep taking the temperature of the water.  I am trying to write my lab report and I have no idea how to calculate the number of calories the wood produced while it was burning.  

[jdurg]

Well, first you have to think about why you kept taking the temperature of the water.  As I'm sure you've discovered, the temperature of the water went up when the wood was burning.  Now why would the temperature go up?  The temperature went up because it absorbed energy from the burning wood.  Calories are a measurement of energy.  If you know how much the temperature changed and you know what the specific heat of water is, you'll know how much energy it absorbed which would then be equal to the energy given off by the wood.  (Well, equal enough for your purposes.  In reality, it's not 100% equal due to various losses).  Also, specific heat is a measurement of how much energy is required to change a specific mass of a substance one degree Kelvin.  (A change of 1 degree Kelvin is equal to a change of 1 degree Celcius.  The units of specific heat are 'Energy/mass*Kelvin'.  So a specific heat of 2.06 J/g*K means that for every 2.06 joules, one gram of the substance will go up in temperature one degree Kelvin).

In order to calculate the calories, you need to find the specific heat of water and the temperature change you observed.  You'll then need to do some basic math and make sure the units match up.  It shouldn't be too difficult.

[AzureIce]

Here's a formula you can work with:

Q = mc?T + mc?T

The first set deals with the can and the second deals with the water.  m is the mass, c, is the specific heat (for water it is 4186, for the can it depends what it is made of - you'll have to look at a table).  ?T is the change in temperature.

Q is the amount of heat in joules, which you can convert directly to calories.

This won't be entirely accurate, of course, because you'll also be heating the air, and anything else that happens to be around, and you can't really calculate it.

[jdurg]

Based upon the initial post, however, the temperature of the can was not recorded so you have to just ignore that portion of the equation.  With the specific heat of most metals being so very low, I don't think that it would play all too much of a role in determining the final answer.  (Also, if accuracy was so very important, they probably would have used a much different setup for performing this experiment).  

[AzureIce]

Based upon the initial post, however, the temperature of the can was not recorded so you have to just ignore that portion of the equation.  With the specific heat of most metals being so very low, I don't think that it would play all too much of a role in determining the final answer.  (Also, if accuracy was so very important, they probably would have used a much different setup for performing this experiment).  

The temperature of the can will be very close to the temperature of the water.  Close enough that they can be considered the same for this given lab.

This lab will already be inaccurate, probably by about 50%.  It isn't a good idea to ignore the can because it is in very close contact, and you *can* calculate it it pretty easily.  Why throw away data if you can use it?

[jdurg]

Well, this is all dependent on whether or not they weighed the can.    If they did not get the mass of the can, then there is no way to calculate the energy it absorbed.  If they have the mass of the can, then yeah, go ahead and figure out the energy it took in.  But if they don't have that data, which I'm assuming they don't, they'll have no choice but to ignore the can aspect.

[AzureIce]

Well, this is all dependent on whether or not they weighed the can.    If they did not get the mass of the can, then there is no way to calculate the energy it absorbed.  If they have the mass of the can, then yeah, go ahead and figure out the energy it took in.  But if they don't have that data, which I'm assuming they don't, they'll have no choice but to ignore the can aspect.

It'd be just silly not to weigh the can, although you are correct.  Can't use the above formula if you don't have all the masses.