Question

Part 1) In the laboratory a student determines the specific heat of a metal as follows: He heats 18.3 grams of lead to 99.28 °C and then drops it into a cup containing 77.7 grams of water at 21.08 °C. When thermal equilibrium is reached, he measures the final temperature to be 21.76 °C. Assuming that all of the heat from the metal is transferred to the water, he calculates the specific heat of lead to be ______ J/g°C.

Part 2) He heats 18.3 grams of copper to 98.40 °C and then drops it into a cup containing 76.5 grams of water at 20.88 °C. When thermal equilibrium is reached, he measures the final temperature to be 22.51 °C. Assuming that all of the heat from the metal is transferred to the water, he calculates the specific heat of copper to be ____ J/g°C

Answer 1

part-1

Heat lose of lead                   =                Heat gain of water

mcT                                  =                 mcT

18.3*c*(99.28-21.76)            =               77.7*4.18*(21.76-21.08)

    c     = 0.156

The specific heat of lead is 0.156J/g-c⁰  

part-2

Heat lose of copper                  =                Heat gain of water

mcT                                  =                 mcT

18.3*c*(98.40-22.51)             =    76.5*4.18*(22.51-20.88)

c   = 0.375

specific heat of copper   = 0.375J/g-c⁰