Do Now (2/16/12) (7-8 min):

1.
2.
3.
4.
5.
Convert the following temperatures:
39˚F=____˚C
10˚C=____K
100 ˚C=____˚F
455 K=____˚C
388K=____˚C
Do Now (2/16/12) (7-8 min):
Convert the following temperatures:
1. 390 K =____˚C
2. 10˚C=____K
3. 0 ˚C=____˚F
Identify the types of heat transfer in each
example:
4. Sitting in the sun
5. Heating up a cup of coffee
6. Heating a pan on an electric stove

Heat Transfer
2/18/11
Heat
The
energy that
is transferred
between objects
Brainstorm (2 min):
Is heat transferred when the temperature of an
object changes?
Brainstorm, and write an answer in your notes.
Discuss with your partner once you’ve
finished.
*thinking point – what does “temperature” mean?
YES!!! Heat transfer means a
change in energy, which means
a change in temperature!!!
Heat Transfer Due to Temperature
Change:
 Heat
lost or gained (Q)
Q  mCT
Q=heat (measured in J)
m=mass
C=specific heat
∆T=Tf –Ti =change in temperature (˚C/K)
Specific Heat ( C )
the amount of heat per unit mass required to
raise the temperature by one degree Celsius;
it is unique to and constant to each material
http://www2.ucdsb.on.ca/tiss/stretton/Database/Specific_Heat_Capacity_Table.html
Measured in:
J
J
J
cal
;
;
;
kg  C kg  K g  C g  C
Specific Heat (C)
Specific
heat can be
looked up in tables
based on the
MATERIAL (ex.
Water = 4180 J/kgK)
Example:
What is the heat gained when a 1.3 kg sample of
water is heated from 10˚C to 30˚C. If the specific
heat of water is 4180 J/kgC, how much heat did the
water gain or lose?
Step 1: List
Step 2: Find an
knowns/unknowns appropriate equation
m=1.3 kg
Ti= 10˚C
Tf= 30˚C
C=4180
Q=?
Q  mCT  mC (T f  Ti )
Step 3: Plug in and
solve
Q  (1.3)( 4180)(30  10)
 108680J
Solving for final and initial
temperatures
*reminder: ΔT=Tf-Ti
 When ask to solve for Tf or Ti, replace ΔT with
Tf -Ti
 Example: Q=mC ΔT  Q=mC(Tf-Ti)

Q
 T f  Ti
mC
Q
 Ti  T f
mC
Heat
If
heat is
absorbed, Q is +
If heat is
released, Q is -
*note

The heat transfer we have discussed today
is for ONE object at a time… tomorrow
we will discuss heat transfer between TWO
objects.
Practice 2/16/12:
 Please
use the rest of class to
work on the paper “Heat
Transfer B.” It is due on Tuesday,
2/21/12
 *for #’s 1 and 2 on your paper,
assume C=4180. Thanks!
 Specific heats: Mercury: 114 J/kgC
 Specific heats are found on p.
279 of your textbook
Do Now (2/17/12):
A 20 kg sample of water is
cooled from 90˚C to
40˚C. If the specific heat
of water is 4180 J/kgC,
how much heat did the
water gain or lose?
Heat Transfer
Heat flows from one object to another
until they are in thermal equilibrium.
 Thermal equilibrium: all objects in a
system have the same temperature

Calorimetry:
science
of
measuring the heat
of physical changes
Heat Transfer

Heat is thermal energy (cannot be
created or destroyed):
Q1  Q2
m1C1 (T f  T1i )  m2C2 (T f  T2i )
Solving For Final Temperature
m1C1T1i  m2C2T2i
Tf 
m1C1  m2C2
Solving for Final Temperature

A 0.4 kg sample of water is at 90˚C is mixed
with a 0.7 kg sample of methanol (specific
heat 2450 J/kgC) initially at 20˚C. Assuming no
heat loss to their surroundings what is the
final temperature of the mixture?
Mouse traps
“Victor”
brand (if you
would like to buy your
own).
You may sign up for me
to buy yours – they are
$1 each.
Practice (2/17/12):
Please use the rest of class to work on
the paper “Heat Transfer.” It is due on
Tuesday!!!
 Specific heats can be found on p. 279 of
your textbook.

Do Now (2/21/12):
A 2 kg sample of water (specific heat
4180 J/kgC) is at 75˚C is mixed with
a 3 kg sample of methanol (specific
heat 2450 J/kgC) initially at 25˚C.
1. List your variables.
2. Assuming no heat loss to their
surroundings what is the final
temperature of the mixture?
Pass in:
Pass in homework (unless you were on
the field trip Thursday); then please take
out the notes you have taken so far in this
unit.
 *note – you will have a short quiz
tomorrow on thermal energy

Do Now:
A physicist plans to change 100 grams (0.100 kg)
of ice at –10.0 oC into 100 g of water at 0 oC.
What is the heat required to raise the ice to
its melting point? The specific heat of ice is
2060 J/kgC
Calorimetry

A 12.9 gram sample of an unknown metal at
26.5°C is placed in a Styrofoam cup
containing 50.0 grams of water at 88.6°C.
The water cools down and the metal warms
up until thermal equilibrium is achieved at
87.1°C. Assuming all the heat lost by the
water is gained by the metal and that the
cup is perfectly insulated, determine the
specific heat capacity of the unknown metal.
The specific heat capacity of water is 4180
J/kg°C
Do Now:

A 30 g sample of an unknown metal at 28°C
is placed in a Styrofoam cup containing 50 g
of water at 89°C. The water cools down and
the metal warms up until thermal
equilibrium is achieved at 87°C. Assuming all
the heat lost by the water is gained by the
metal and that the cup is perfectly insulated,
determine the specific heat capacity of the
unknown metal. The specific heat capacity of
water is 4180 J/kg°C
Do Now:
A 0.4 kg sample of water (specific heat 4180
J/kgC) is at 90˚C is mixed with a 0.7 kg
sample of methanol (specific heat 2450
J/kgC) initially at 20˚C. Assuming no heat
loss to their surroundings what is the final
temperature of the mixture?
Calorimetry
Calorimetry : science of measuring the
heat of physical changes
 Calorimeter – a device used for
measuring specific heat capacity
