Specific Heat Capacity
Or, What Does It Take To Heat Up or Cool Down??
HOT SURFACES AND HOT MATERIALS – BE CAREFUL!
Conservation of energy is an extremely important aspect of physics and it may be demonstrated in many different ways. This laboratory uses the concept of Conservation of Heat Energy along with that of Specific Heat Capacity in order to determine the type of metal of which the cylinder is composed.
Fill the 600 mL beaker with approximately 300 mL of water. Adjust the hotplate temperature so that the water in the beaker will reach a temperature of 850 – 900 C. Carefully mass the cylinder at your station and record this in a data table as mcyl. Measure 100 mL of distilled water into a flask and transfer this water to the styrofoam calorimeter. Since 1mL of pure water has a mass of 1 gram, you now have 100 grams of water in the calorimeter. Record the initial temperature To (cal) of the calorimeter water.
Using the tongs, carefully immerse the cylinder into the water bath on the hot plate. After the cylinder has been immersed for approximately 5 – 6 minutes, record the temperature of the water bath. This temperature will also be the temperature of the metal cylinder T0 (cyl). Again using the tongs, carefully remove the cylinder from the water bath and quickly place it into the calorimeter.
After several minutes, you will observe that the temperature has risen. Record in your data table the highest temperature attained by the calorimeter water. It is very important that the temperature of the calorimeter water is not taken until it has reached the highest point. This final calorimeter temperature Tf (cal) is also the final temperature of the metal cylinder Tf (cyl). Also record the difference between the final temperature and the initial temperature of both the metal cylinder and the calorimeter water.
Assuming that all of the heat acquired by the metal cylinder while in the water bath was effectively transferred to the calorimeter, it is now possible to calculate the specific heat of the metal cylinder.
? H of the calorimeter water = ? H of the metal cylinder or
Heat gained by the calorimeter water is equal to the heat lost by the cylinder.
Since H = (m) (cp) (?T),
(cp water) (mass water) (?T water) = (cp cyl) (mass cyl) (?T cyl)
Simply manipulate the above equation to solve for the Specific Heat Capacity
of the cylinder:
cp cyl = (Cp water) (mass water) (?T water)
(mass cyl) (?T cyl)
1) Explain why / how the temperature of the calorimeter water increases and the temperature of the metal cylinder decreases.
2) Calculate the amount of heat that was transferred from the metal cylinder to the calorimeter water.
3) Of what metal is your cylinder composed and on what do you base your answer?
4) What percent error is associated with your answer to number 3 above?
This lab write-up shall consist of the following:
Answers to Questions
Analysis & Conclusion