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Physics 105 lab 2 answer key with data
Typology: Lab Reports
Uploaded on 02/16/2022
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Goal. Determine the specific heat of two metals by heat transfer. Background. A review of the physics of heat exchange can be found in your textbook. The specific heat c is a measure for how much heat 1 gram of a material can absorb until its temperature increases by 1 degree. If an object of mass M absorbs heat Q then its initial temperature T i increases to a final temperature T f according to the equation Q = M c ( T f – T i) (1) If a heated metal sample (mass M metal) of initial temperature T i,metal is placed into cold water (mass M water) with temperature T i,water, the metal will heat the water until both have the same final temperature T f,both. Since the heat gained by the water Q water and the heat lost by the metal – Q metal are identical, Q water = – Q metal, we have: M water c water ( T f,both – T i,water) = – M metal c metal ( T f,both – T i,metal) (2) This equation can be solved for the specific heat cm for the metal: (3) The values of the specific heats for the materials used in this experiment are given below: Material Specific heat c metal (cal / gC) water 1. aluminum 0. copper 0. Measurements Task 1. Prepare the water and the aluminum sample for heat transfer. Determine the mass of the (cold) water M water in the cup and its temperature T i,water. Also determine the mass of the aluminum sample M metal and heat it in boiling water T i,metal. I measured MAl = 83.4g I measured MCu = 255.7 g Ti,metal = 99.1 0 C Troom = 20.1 0 C Task 2. Carry out the heat transfer. After the temperature of the aluminum sample T i,metal has stabilized after a few
Cu Trial 1 C^1 Cu=-366.0(20.2-15.4)/255.7(20.2-99.1)=0.087cal/gC |0.087-0.092/0.092|100%=5.43% Cu Trial 2 C^2 Cu=-265.0(23.9-16.9)/255.7(23.9-99.1)=0.096cal/gC |0.096-0.092/0.092|100%=4.35% Task 5. Lab report Suppose the original temperature T i,water of the water was far below room temperature. Would the specific heat determined by (3) be larger or smaller? Is (3) still valid in this case? The specific heat would be smaller because some heat will be released. It is still a valid case, the ratio becomes lesser value than the previous value. Include your data sheet and answers to the above questions in your report