[DonalB]

Hi, I have a couple of questions concerning my science fair project.

I built a bomb calorimeter for my science fair project to run tests on the beef.
Aside from that, How do I find its heat capacity. I am aware that I need to use a substance that I can find information about and that is combustible. I chose Isopropyl alcohol, C₃H₈O. The alcohols specific heat is 2.56J g^-1K^-1.

If I burn 10g of the substance in my calorimeter I might a change in temperature of 30 degrees^oK with aprox. 3785 g of water. How do I find the heat capacity of the calorimeter with this data.

I havnt carried out the actually combustion of the substance yet, I just threw out some numbers.

So how do I use this data to find the heat capacity of the calorimeter?


Thanks,
Donal.

[Alpha-Omega]

EXAMPLE

4. A. Estimating Cv

The heat capacity of the bomb calorimeter can be estimated by considering the calorimeter to be composed of 450 g water and 750 g stainless steel. Knowing the specific heat capacity of water to be 1 cal/g·K and estimating the specific heat capacity of steel to be 0.1 cal/g·K yields

See Picture


Good Luck at the Science Fair...I have judged quite a few amazing things there....

[DonalB]

Ty,
however, Isn't the heat of the substance combusted also needed to find the heat capacity?

or maybe im just confused

[Alpha-Omega]

OK sorry then you asked for the Cv of the calorimeter....

[DonalB]

I am interested in the heat capacity of the Calorimeter, what does C_v mean? Im not familiar with the variable you are using.

Thanks,
Donal.

[Kryolith]

To determine the heat capacity of a calorimeter, a good way is to mix warm water with cold water.
If you add warm water (mass m₂, temperature T₂) to cold water (m₁, T₁) in a calorimeter a heat exchange occurs until the water has the mix temperature T_m.
ΔQ₂ = m₂*c_w*(T₂ −T_m)
The delivered heat ΔQ₂ is absorbed by water AND the calorimeter, so try to find out the equation for ΔQ₁.

Assumption: perfect adiabatic process, in practice you have to consider the heat exchange with the environment.

[Alpha-Omega]

OH I see what you are asking...hang on minThe heat capcitiy for the Calorimeter is denoted by Cv....I sent you an example and you do need it for your experiment...OK then ...determination of the heat capacities for your samples next....

[Alpha-Omega]

You can determine the heat capacity of the system by adding a known amount of electrical energy and measuring the temperature increase.

Alternatively, you can burn a compound whose heat of combustion is known (a standard) and from the measured increase in temperature determine the heat capacity of the system. I

n most experimental situations for conventional bomb calorimetry, the heat capacities of reactants and products are small compared with the heat capacities of water and metal in the calorimeter, itself. In this experiment, for example, less than 1 g of compound is burned and approximately 2 L (~ 2 kg) of water plus the metal calorimeter are heated.

For relatively small changes in temperature (a few degrees), the heat capacity of the system can be considered constant. Consequently, one can determine the heat capacity of the system from the following equation:

I think this might help you:  http://www.chem.ufl.edu/~itl/2045/lectures/lec_9.html

[Borek]

The heat capcitiy for the Calorimeter is denoted by Cv...

You are doing everything that can be done and then some to confuse people. Cv is usually reserved for specific heat capacity at constant volume, as opposed to Cp - specific heat capacity at constant pressure. Why do you use this symbol for calorimeter heat capacity?

[Alpha-Omega]

Why are you confused?  I do not see the confusion...I can be flexible....NO SUBSCRIPT DESIGNATIONS....I can remove Cv and Cp...and set it up as follows:


OK so do it this way:

Calorimetry

Heat Capacity of the Calorimeter:

In calorimetry it is often desirable to know the heat capacity of the calorimeter itself rather than the heat capacity of the entire calorimeter system (calorimeter and water). The heat (q) released by a reaction or process is absorbed by the calorimeter and any substances in the calorimeter. If the only other substance in the calorimeter is water, the following energy balance exists:


q = qcal + qw


where qcal is the heat flow for the calorimeter and qw is the heat flow for the water.

Both of these individual heat flows can be related to the heat capacity and temperature change for the substance.


qcal = Ccal ΔT

qw = Cw ΔT

where Ccal is the heat capacity of the calorimeter and Cw is the heat capacity of the water. Because the water and calorimeter are in thermal equilibrium, they both have the same temperature and thus ΔT is the same for both. The consequence is that the heat capacity of the entire system (C) is the sum of the heat capacities for the individual components.


C = Ccal + Cw

The heat capacity is an extensive property; that is, the heat capacity depends upon the amount of substance present. The calorimeter exists as a fixed unit, thus its heat capacity is a fixed value. The amount of water in the calorimeter, however, can vary, and thus the heat capacity of the water can vary. When dealing with variable amounts of material, one often prefers to use an intensive measure of the heat capacity. One common intensive version of the heat capacity is the specific heat capacity (s), which is the heat capacity of one gram of a substance.


s =  Cw

mw


Because the mass of water (mw) and the specific heat capacity of water are both known, you can readily calculate the heat capacity of the water. The specific heat capacity of water (sw) is


sw = 4.184 J oC-1 g-1

Overall one can write


C = Ccal + sw mw