Specific Heat of Copper - Bibb County Schools

Thermal Capacity



Of a particular body is the energy required to raise the temperature of that body by 1°C.



Thermal capacity =

change in thermal energy

temperature change



C = ∆Q / ∆T


Example 1



A 2 kg cylinder of copper is heated from room temperature (20˚C) to 500ºC. 374kJ of thermal energy
were transferred to the copper during the heating process. Calculate the thermal capacity of this piece
of copper.



Answer: 780J


Example 2



A 25kg cylinder
of copper is heated from room temerature. The same 374kJ of thermal energy were
used during the heating process but this time the copper’s temperature rose from room temperature
to only 58.4ºC. Calculate the heat capacity of this piece of copper.



Answer:

9740J What is the difference between Ex 1 and Ex2


Specific Heat



Adding energy to a material causes the temperature to go up.



Taking energy away from a substance causes the temp. to go down!



Have you ever noticed that on a hot summer day the pool is coo
ler than the hot cement?

OR maybe
that the ocean is cooler than the hot sand?

Why? The sun has been beating down on both of them for
the same amount of time...........



It takes more thermal energy to raise the temperature of water

that it does the cement!


Specific Heat



The amount of energy required to raise the temperature of a material (substance).



It takes different amts of energy to make the same temp change in different substances.



The specific heat capacity of a particular substance is equal to the
energy required to raise the
temperature of a 1kg mass of the substance by 1ºC


Specific Heat of water



The Cp is high because H
2
O mols. form strong bonds w/each other.



It takes a lot of energy to break the bonds so that the the

molecules can then start to move around
faster (HEAT UP).



Example:

Specific Heat of Water



Cp = 4,184 Joules of energy to raise the temperature of 1kg 1°C.

Example 3



A 2kg cylinder of copper is heated from room temperature(20ºC) to 500ºC. 374kJ of thermal

energy
were transferred to the copper during the heating process. Calculate the specific heat capacity of this
piece of copper.



Answer: 390J


Example 3



A 2kg cylinder of copper is heated from room temperature(20ºC) to 500ºC. 374kJ of thermal energy
were
transferred to the copper during the heating process. Calculate the specific heat capacity of this
piece of copper.



Answer: 390J


Example 4



A 25kg cylinder of copper is heated from room temerature. The same 374kJ of thermal energy were
used during the he
ating process but this time the copper’s temperature rose from room temperature
to only 58.4ºC. Calculate the specific heat capacity of this piece of copper.



Answer: 390J


Solving for specific heat



There are two methods common for measuring the specific h
eat capacity.



Electrical
–

If an electrical immersion heater is place into a solid or a liquid, then the energy from
the heater will be transmitted by conduction into the substance and the substance will get hotter.



Mixtures
–

If a hot object is place next

to a cooler one (or placed into it if the cooler one is liquid),
then the cooler substance will gain energy and become hotter and the hotter object will lose
energy and become cooler until both objects come to the same temperature called thermal
equilibri
um.


Example 5
–

electrical



A 240V electric heating element is used to heat water. The temperature of the water rose from 20ºC to
50ºC in 4minutes 20 seconds. During the heating process, the current flowing in the heater was
measured to be 3.54A. Calcula
te the mass of the water.

Solution



First the power rating is Power = Voltage x Current



P = V I (I.B. Data booklet page 7)



P = 240 x 3.54 = 850W



The heater supplies 850J of energy to the water every second (850W= 850J/s). So in 4minutes
20seconds(260s),
energy transferred to the water = 850 x 260 = 221x10
3
J.



Answer: 1.75kg


Example 6



The 850W heater was then placed into a hole in a piece of copper of mass 1.75kg. (A) Calculate the
temperature rise in the copper if the heater was left on for 4min 20sec.
(B) Calculate the final
temperature of the copper if the heater was left on for 10min and the copper was originally at a
temperature of 65ºC.



Answer: (A) = 324ºC, (B) = 747ºC


Example 7
–

Mixture



A block of substance “X” has a mass of 100g and is heated to

260ºC. The block is then placed into a
beaker containing 500g of water at 20ºC. After some time both substances reach their equilibrium
temperature of 30ºC. Calculate the specific heat capacity of substance X.



Solution: Energy gained by the water = Energ
y lost by X



Q
w

= Q
x



m
w
c
w
∆
T
w

= m
x
c
x
∆
T
x




Answer = c
x

= 913J/kgºC


Example 8



How much energy is needed to heat a 1kg aluminum pan containing 2kg of water from 25ºC to 95ºC?



Solution: Total Energy = Energy gained by aluminum + Energy gained by water



Answer 6
51kJ




Example 9



A 0.5kg block of copper (specific heat capacity 390J/kgºC) at an initial temperature of 420ºC was
placed into 1.3kg of water at 40ºC. What will be the final temperature of the mixture when thermal
equilibrium is reached?



Answer: T
final

=

53.1ºC




Micro Properties of different phases



Solids



Strong bonds between atoms



Lowest internal energy



Atoms in fixed positions vibrating/oscillating



Liquids



Weaker forces. Some bonds are broken



More internal energy



Atoms can move about and change places



Gases



Virtually no forces/bonds



High internal energy



Atoms completely free to move at high speed



Macro Properties of different phases



Solids



Maintain shape



Lowest temp



Low compression/expansion



Liquids



Takes the shape of its container



Moderate temp



Low
compression/expansion



Gases



Fills the container



Highest temp



High compression/expansion



Plasmas
–

atoms are at extremely high temperatures and are ionized.



Usually found in stars.


















Substance

Specific Heat(J/kg.
°C)

Specific Heat of
Beryllium

1830

Specific Heat of Cadmium

230

Specific Heat of Copper

387

Specific Heat of Germanium

322

Specific Heat of Gold

129

Specific Heat of Iron

448

Specific Heat of Lead

128

Specific Heat of Silicon

703

Specific Heat of Silver

234

Specific Heat of Brass

380

Specific Heat of Glass

837

Specific Heat of Ice(
-
5°C)

2090

Specific Heat of Marble

860

Specific Heat of Wood

1700

Specific Heat of Alcohol(ethyl)

2400

Specific Heat of Mercury

140

Specific Heat of Water(15°C)

4186

Specific Heat of Steam(100°C)

2010

Specific Heat of Aluminium

900

Specific Heat of Tin

540

Specific Heat of Steel

120

Specific Heat of Sand

830

Specific Heat of Ethanol (Alcohol, ethyl
32°F)

2.3 K