Heat Transfer
Q = mcΔT
Factors in Heat Transfer
1.
•
•
Temperature
a large change in temperature indicates a
large energy change
Symbol: ΔT
2. Mass of Substance
• The greater the mass of substance, the
greater the amount of heat it can absorb
and release
• Symbol: m
3. Type of Substance
• Specific heat capacity is the quantity
of energy, in Joules (J) that is
required to change one gram (g) of the
substance by one degree Celsius (°C)
• The specific heat capacity reflects
how well the substance can store
energy
• A substance with high SHC can absorb
and release more energy
• Symbol: c (units J/g•°C)
To calculate Heat Transfer
Specific heat capacity
(J/g•°C)
Heat transfer
(J)
Q =
m
Mass
(g)
x
c x
ΔT
Change in tempera
(°C or K)
60.0 g of water went from 26.5°C to
9.7°C. Calculate how much heat
was lost by the water.
Given:
M = 60.0 g
ΔT = 9.7 – 26.5 = -16.8°C
Cwater= 4.184 J/g°C (from page 595 Text)
Q= m x c x ΔT
= (60.0g)(4.184)(-16.8°C )
= -4217.472 J
= - 4.22kJ
The heat change of water was
-4.22kJ.
A sample of vegetable oil with a mass
of 0.0600 kg went from 35.0°C to
5.2°C and lost 4.0 kJ of heat.
Calculate its specific heat capacity.
• c= Q =
-4.0 x 103 J
mΔT
(60.0g)(5.2 -35.0°C)
= 2.2 J / g°C
The specific heat capacity of oil is 2.2 J / g°C
Do Practice Problems 8, 10, 11 bc, 12 – 15 (p.
597 – 599) # 3a, 4 – 7 p. 600
Calorimetry
A calorimeter is used to measure changes
of temperature in chemical reactions
If the temperature increases, the
reaction is exothermic.
If the temperature decreases, the
reaction is endothermic.
The heat produced or absorbed by the
reaction is equal to the heat lost or
gained by the surrounding water.
Heat lost = Heat gained
- Qlost
= + Qgained
A piece of metal with a mass of 140 g is heated
to 95.0°C. When the metal is placed in 24.7 g of
water at 22.0°C, the temperature of the water
rises to 47.0°C. Calculate the specific heat
capacity of the metal and identify it.
Given – Water
Remember:
m = 24.7 g
Heat lost = Heat gained
TI = 22.0 °C
- Qlost = + Qgained
TF = 47.0 °C
c = 4.184J/g °C
Qwater= mcΔT
=24.7(4.184)(47.0 – 22.0)
= 2583.62J
Given – metal
m = 140 g
TI = 95.0 °C
TF = 47.0 °C
Qmetal= -2583.62J
C = Q = -2583.62J
mΔT (140g)(47.0 – 95.0)
= 0.384 J/g °C
To find out what the metal is, look at the
specific heat capacity chart on page
595. T
This metal is copper.
For calorimeter calculations, we
assume:
1. The calorimeter is isolated.
2. There is no heat lost or gained from
the environment.
3. Water always maintains its properties,
even when something else is placed in
it.
Try:
p. 607-608 # 18, 19,20, 21
p. 618 # 2, 4, 7
p. 627 – 628 # 10, 16, 22