10 3 part a notesfilled in

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Chp. 10.3 (part A) Notes

1. Specific Heat                                 beakers at the same temperature are
a. Some things ____heat__ up or               supplied with the same amount of
___cool_____ down faster than others.            energy, which will have the higher final
b. Specific heat is the __amount_ of          temperature? Ethyl Alcohol
heat required to raise the temperature of
1 kg of a material by one degree (C or           3. A calorimeter is used to help
K).                                              measure the specific heat of a
substance.

4. Specific heat capacity equation:
The formula is __Q = mc∆T.
Cp = specific heat capacity at constant
pressure
Q = energy transferred as heat
m= mass
∆T= change in temperature

5. Place the variable in the triangle
Specific       How long                   below and write the equation for each
Heat           it takes to heat up        variable.
664 J/kg˚C     fast                                        Q
Land

m x      c       x ∆T
4186 J/ kg˚C     slow
Water

2. Write a summary statement based on
the information in the chart about how
specific heat value determines how long
a substance takes to heat up or cool
down. The lower the specific heat
value, the quicker it heats up. The              A) The temperature change in ºC is
higher the specific heat value the               equal in magnitude to a 1K, so _NO____
slowe it heats up.                               conversion is necessary.
Concept Question:
Ethyl alcohol has about one-half the
specific heat capacity of water. If equal
masses of alcohol and water in separate
Examples                                         4. Finding the final temperature with
a) How much energy must be                       two substances
transferred as heat to 420 kg of water in           Energy absorbed by water = energy
a bathtub in order to raise the water’s                 released by the substance
temperature from 25°C to 37ºC?                                  Qw = -Qx
Specific Heat of Water = 4186 J/kg•ºC                     C1m1∆T1 = C2m2∆T2
m=420                   Q=mc∆T
∆T=37-25=12        Q=420(12)4186                  ∆T should           Tf= c1m1ti1 + c2m2ti2
c=4186             Q = 2.11 x 107J                always                   c1m1 + c2m2
be positive.
b) What mass of water is required to
absorb 4.7 x 105 J of energy from a car          Teacher Example:
engine while the temperature increases           1. What is the final temperature when a
from 298K to 355K? Specific Heat of              3.0kg gold bar at 99˚C is dropped into
Water = 4186 J/kg•ºC                             0.22kg of water at 25˚C?
Q=4.7 x 105           m=Q/(c∆T)                  m1=3        Tf= c1m1ti1 + c2m2ti2
∆T=355-298=57         m=4.7 x 105                T1=99
c=4186                    (4186 x 57)            m2=0.22           c1m1 + c2m2
m=1.97kg                     T2=25     = (129x3x99 + 4186x0.22x25)
Student Practice                                 c1=129       (129x3 + 4186x.22)
a) A vanadium bolt gives up 1124J of             c2=4186 = 46.9˚C
energy as its temperature drops 25K. If
the bolt’s mass is 0.093kg, what is its          2. A 0.225kg sample of tin initially at
specific heat?                                   97.5ºC is dropped into 0.115kg of water
Given Formula Substitution Answer                initially at 10.0ºC. If the specific heat
Q=1124            c=Q/(∆Tm)                      capacity of tin is 230 J/kgºC, what is the
∆T=25            c=1124       =483J/kg˚C         final temperature of the tin-water
m=0.093          (25 x 0.093)                    mixture?
m1=0.225          Tf= c1m1ti1 + c2m2ti2
b) How much energy must be added to a            T1=97.5
bowl of 125 popcorn kernels in order for         m2=0.115           c1m1 + c2m2
them to reach a popping temperature of           T2=10
175ºC? Assume that their initial                 c1=230
temperature is 21ºC, that the specific            c2=4186
heat capacity of popcorn is 1650J/kgºC,          =(230x0.225x97.5+4186x0.115x10)
and that each kernel has a mass of                         (230x0.225 + 4186x.115)
0.105g.                                          = 18.5˚C
∆T=175-21=154 Q=mc˚∆T
c=1650               =0.013(1650)154
m=0.013              = 3303J
To determine mass you need to first convert
0.105g to kg by dividing by a 100. You will
get 0.000105kg. Then you multiply that
0.000105kg by 125 to determine the total
mass.
Student Practice
3. What is the final temperature when
0.032kg of milk at 11ºC is added to
0.16kg of coffee at 91ºC? Assume the
specific heat capacities of the two liquids
are the same as water, and disregard any
energy transfer to the liquids’
surroundings.
m1=0.032         Tf= c1m1ti1 + c2m2ti2
T1=11
m2=0.16           c1m1 + c2m2
T2=91
c1=4186
c2=4186
=(4186x0.032x11+4186x0.16x91)
(4186x0.032 + 4186x.16)
= 77.7˚C

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