capacity of the thermometer, which records the temperature, can be neglected. Let the final temperature of the mixture be T3. Energy conservation gives:Qlost(specimen) = Qgained(Water) + Qgained(Calorimeter)which yields the unknown specific heat c1 of the specimen asc1 = (m2c2+m3c3) (T3 - T2) m1(T1 ? T3) We assume that the mixing can be done without loss of heat by the hot specimen to the surroundings.We will consider a specimen heated to a high temperature is dropped into water contained in a calorimeter cup at a lower temperature. If this system is thermally insulated from the surroundings, the specific heat of the specimen can be determined by equating the heat lost by the metal to the heat gained by both the calorimeter cup and the water contained in it. (http://www.physics1.howard.edu/MSIP/GenLab1/GL1-10.pdf )III.Diagram/MaterialsIV.Procedure The initial masses of boiler dipper, copper, lead, dipper and shots, inner vessel with water, and water are measured and recorded. The specimens (copper/lead) are heated and the initial temperatures are recorded after which, it is dropped in the calorimeter containing cold water. The temperature rise of the water in the calorimeter is observed and recorded. From the data gathered, the specific heat of the specimen can be known through the formula stated above.V.DataMaterial of SpecimenLeadCopperMass of Boiler Dipper, m1(gm)8585Mass of Dipper and Shots,m2 (gm)320280Mass of shots,ms=m2-m1 (gm)235195Mass of inner vessel w/ water,m3 (gm)220220Mass of water, mw=m3-mc (gm)165165Temperature of hot shots,T1 (oC)8384Initial Temperature of System, T2 (oC) 2221Final temperature of system,T3 (oC)2626Specific heat of water,cw (cal/gmoC)1.001.00Specific heat of specimen,cw (cal/gmoC)(Experimental)0.0520.072Specific heat of specimen (cal/gmoc) (standard)0.0360.092Percentage Error 44.4%21.7%VI.Sample Computationc= Q/m?T Qlost(specimen) = Qgained(Water) + Qgained(Calorimeter) ...
