Part A In the following experiment, a coffee-cup calorimeter containing 100 mL o
ID: 969113 • Letter: P
Question
Part A
In the following experiment, a coffee-cup calorimeter containing 100 mL of H2O is used. The initial temperature of the calorimeter is 23.0 C. If 8.00 g of CaCl2 is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? The heat of solution Hsoln of CaCl2 is 82.8 kJ/mol.
Express your answer with the appropriate units.
Part A
It takes 46.0 J to raise the temperature of an 10.0 g piece of unknown metal from 13.0C to 24.9 C. What is the specific heat for the metal?
Express your answer with the appropriate units.
???
Parts B and C
The next two questions pertain to silver. They have nothing to do with unknown metal described in Part A.
Part B
The molar heat capacity of silver is 25.35 J/molC. How much energy would it take to raise the temperature of 10.0 g of silver by 17.7 C?
Express your answer with the appropriate units.
Part C
What is the specific heat of silver?
Express your answer with the appropriate units.
Value / units = ???
Value / Units???Explanation / Answer
Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g)
c = specific heat capacity (4.19 for H2O) in J/(g oC)
T = change in temperature = Tfinal - Tinitial in oC
Part A
8 g CaCl2 x (1 mol / 111 g) = 0.072 mol
0.072mol CaCl2 x (82.8 kJ / mol) x (1000 J / kJ) = -5967 J
Assume that the heat capacity of the solution is the same as that of water (4.184 J/gC)
heat = mass x specific heat x T
mass = 100 gm H2O + 8gm CaCl2
specific heat = 4.184
T = heat divided by (mass x specific heat)
T = -5967 J divided by (108 x 4.184) = -13.2 C
Final temp should be 23 + 13.2 =36.2 C
Part A
Q = m c T
Q = 46
m = 10
c= ?
T = 24.9-13 = 11.9
Substitute in the equation
46 = 10 x c x 11.9
c = 0.3865
specific heat for the metal is 0.3865
PART B
specific heat of silver
25.35 J/molC x (1 mol / 107.9 g) = 0.2349 J/gC --- {Where 107.9 is the molar mass}
specific heat of silver = 0.2349 J/gC
Q = m c T
Q = ?
m = 10
c= 0.2349
T = 17.7
Q = 10 x 0.2349 x 17.7 = 172.84 J is required to raise the temperature by 17.7 C