Question

for physics heat capacity lab writeup from a calorimeter experiment, we are asked to show the algebra behind calcupaltions for Cp in cal/gdegreeC in the case wehere the cup and shot are the same metal??? The cup is composed of Aluminum and then the alumum metal used for metal shot samples was 50.5 grams. the water was 150.4 grams and with ice added became 178.4 grams and the tempinital was 22 C and final 11 degrees C and the shot Ti was 85 C and final 16.4 degrees C

Answer 1

Given
   mass of cup and shot is m c = 50.5 grams made up of same material aluminium

  

   mass of water=150.4 g and ice mi=28 g both are at Ti =22 0C, Tf = 11 0C

   and the temperature of cup and shot , ti = 85 0C, tf = 16.4 0C

here ice will melt at 0C and temperature will increase from 0 to 11 0C

   mi*lf + mi*c*dT+mw*c*dT = m_C*C_C*dT

   0.028*334*10^2 + 0.028*4186*(11)+0.1504*4186(11) = 0.0505 *C_C*(85-16.4)

       C_C = 2641.17 J/kg 0C
specific heat of given cup and shot is 2641.17 j/kg 0C