Download Notes - hrsbstaff.ednet.ns.ca

Heat Calorimetry
q  mcT
q
ΔT = Tfinal – Tinitial
m
c
Example 1. Solving for q
10.0 g of ice was added 60.0 g of water. The initial temperature of the
water was 26.5oC. The final temperature of the mixture was 9.7oC. What amount
of heat was lost by the water?
We are going to use the mass of the water (60.0 grams).
The Tf is 9.7oC, the Ti is 26.5oC
Now we use the formula:
q  mcT

J 
q  (60.0 g ) 4.184 o (9.7 o C  26.5 o C )
g C

q  4217.5 J
Example 2. Solving for specific heat capacity (c)
Calculate the specific heat capacity of peanut oil if the mass is 65.0 grams,
the initial temperature was 35.0oC and the final temperature was 5.2oC and the
amount of heat was - 4000J (negative because it lost heat)
Needing to find “c”, we must rearrange the equation to solve for c.
q  mcT
Divide both sides by mΔT to solve for c.
q
c
mT
q
c
mT
 4000 J
c
(65.0 g )(5.2 o C  35.0 o C )
c  2.07
J
g oC
Example 3. Solving for temperature (∆T then final temperature)
ΔT
A 150g sample of water at 25oC is heated until it has absorbed 2200J of
heat. What is the final temperature of the water?
Q  mcT
Q
T 
mc
Divide both sides by mc to solve for ∆T.
T 
2200 J
(4.18 J o )(150 g )
g C
T  3.51o C
Remember that ∆T = Tf-Ti so,
3.51oC = Tf - 25oC
Tf = 3.51oC + 25oC
Tf = 28.51oC
Example 4. Solving for mass (m)
A sample of mercury (c = 0.14 J/goC) is heated from 25.5oC to 52.5oC. In
the process 3050J of heat are absorbed. What mass of mercury was contained in
the sample?
q  mcT
m
Divide both sides by c∆T to solve for m.
q
cT
3050 J
(0.14 J / g C )(52.5 o C  25.5 o C )
m  806.9 g
m
o
Practice Problems:
1. A sample of an iron nail (c = 0.444 J/goC) absorbs 18.2J of heat as it is
heated from 23.5oC to 35.2oC. What is the mass of the nail?
(Answer: m = 3.5g)
2. A 33.7 g silver spoon is put into a cup of hot coffee. It takes 0.435 kJ of
energy to change the temperature of the spoon from 22.5oC to 84.5oC.
What is the specific heat capacity of the silver spoon?
(Answer: c =0.208 J/goC)
3. How much heat is released by 55.7 g of gold (c = 0.129 J/goC) when its
temperature is increased 20oC?
(Answer: Q = 143.7J)
4. Determine the final temperature when 45g of aluminum (c=0.900J/goC) is
cooled from 65oC releasing 525J of heat (release = negative).
(Answer: ∆T= -12.96oC, Tf =52.04oC)