Download Specific Heat Capacity

UNIT 3: Energy
Specific Heat Capacity
Specific Heat Capacity (C or S ) - The quantity of heat required to raise the temperature of a
substance by one degree Celsius is called the specific heat capacity of the substance. The quantity
of heat is frequently measured in units of Joules(J). Another property, the specific heat, is the heat
capacity of the substance per gram of the substance. The specific heat of water is 4.18 J/g° C.
Substance
Air
Aluminum
Copper
Gold
Iron
Mercury
NaCl
Ice
Water
C (J/g oC)
1.01
0.902
0.385
0.129
0.450
0.140
0.864
2.03
4.18
q = m x C x T
q = m x C x (Tf - Ti)
q = amount of heat energy gained or lost by substance
m = mass of sample
C = heat capacity (J oC-1 g-1 or J K-1 g-1)
Tf = final temperature
Ti = initial temperature
Solving for Specific Heat of a metal dropped in water:
A 245.7g sample of metal at 75.2 degrees Celsius was placed in 115.43g water at 22.6 degrees Celsius. The
final temperature of the water and metal was 34.6 Celsius. If no heat was lost to the surroundings what is
the specific heat of the metal?
-qmetal=qwater
-(mCT)=mCT
-(mC(Tf-Ti))= mC(Tf-Ti)
o
- (245.7g x C x (34.6 C-75.2oC))= 115.43g(4.18J/goC)(34.6oC-22.6oC)
C x (9975goC)=5790J
0.580J/goC =C
Solving for the Final Temperature when metal is dropped in water:
Determine the final temperature when a 25.0g piece of iron at 85.0°C is placed into 75.0grams of water at
20.0°C. The specific heat of iron is 0.450 J/g°C. The specific heat of water is 4.18 J/g°C.
-qmetal=qwater
-(mCT)=mCT
-(mC(Tf-Ti))= mC(Tf-Ti)
-(25.0g(0.450J/goC)(Tf-85.0oC))=75.0g(4.18J/goC)(Tf-20.0oC)
956.25-11.25Tf=313.5Tf-6270
7226.25=324.75Tf
7226.25/324.75=Tf
22.3oC=Tf
Solving for Final Temperature when ice is added to water:
What is the final temperature after a 21.5 gram piece of ice at 0 is placed into a Styrofoam cup with 125.0
grams of water initially at 76.5oC? Assume no loss or gain of heat from the surroundings.
energy to melt the ice
q=mHf
q=21.5 x 334j/g=7181J
energy to bring the water to 0oC
vs.
q=mCT
q=125.0g (4.18J/goC)(76.5oC) =39,971J
The ice will melt, so the letft over energy is
39,971J - 7181J= 32,790J
Reapply to q=mCT, combine the mass of ice and water, assuming we are at a temp. of 0 oC.
32,790J= 1146.5g(4.18j/gC)(Tf-0)
32,790J/(1146.5g(4.18j/gC))=Tf
53.5oC=Tf