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					<press any key to continue>
 How much heat is required to raise the temperature of 68
 g of aluminum fluoride from 25oC to 80oC?



Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.




                      <press any key to continue>
     How much heat is required to raise the temperature of 68
     g of aluminum fluoride from 25oC to 80oC?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
           HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
J/goC.
So such a problem has 4 variables -- Heat, Mass, T, Cp
Looking at our problem, we see the following values for these
variables……..
Heat = x
Mass = 68 g
T = 80o C - 25o C = 55o C
Cp   = ?
                           <press any key to continue>
   How much heat is required to raise the temperature of 68
   g of aluminum fluoride from 25oC to 80oC?
          HEATgained or lost = (Mass)(T)(Cp)
                        Heat = x
                        Mass = 68 g
                        T = 55o C
                        Cp     = ? = 0.895 J/goC
 Since the problem does not give us the Cp for aluminum fluoride,
 we refer to our list of Common Specific Heats and find that the
 value is………...          Cp AlF3 = 0.895 J/goC
We can substitute all of our values into the equation and solve the
problem……...
              x = (68 g)(55oC)(0.895 J/goC)
              x = 3347.3 J



                          <press any key to continue>
 In order to make 4 cups of tea, 1000 g of water is heated
 from 22o C to 99o C. How much energy is required?



Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.




                       <press any key to continue>
     In order to make 4 cups of tea, 1000 g of water is heated
     from 22o C to 99o C. How much energy is required?
Whenever we have a problem which deals with HEAT, we
must recall the following relationship….
           HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
J/goC.
So such a problem has 4 variables -- Heat, Mass, T, Cp
Looking at our problem, we see the following values for these
variables……..
Heat = x
Mass = 1000 g
T = 99o C - 22o C = 77o C
Cp   = ?
                           <press any key to continue>
   In order to make 4 cups of tea, 1000 g of water is heated
   from 22o C to 99o C. How much energy is required?
          HEATgained or lost = (Mass)(T)(Cp)
                        Heat = x
                        Mass = 1000 g
                        T = 77o C
                        Cp     = ? = 1.0 cal/goC
 Since the problem does not give us the Cp for water, we refer to
 our list of Common Specific Heats and find that the value
 is………...              Cp H2O = 1.0 cal/goC
We can substitute all of our values into the equation and solve the
problem……...
              x = (1000 g)(77oC)(1.0 cal/goC)
              x = 77,000 cal



                          <press any key to continue>
 If a piece of gold with a mass of 45.5 g and a
 temperature of 80.5oC is dropped into 192 g of water at
 15oC, what is the final temperature of the system?


Before continuing on with this tutorial, you should
copy down the above problem and do the best you can
do to solve it.
Once you have the calculation completed, you should
then go through the tutorial and see how you
performed, paying particular attention to any areas
where you may have made a mistake.




                      <press any key to continue>
     If a piece of gold with a mass of 45.5 g and a
     temperature of 80.5oC is dropped into 192 g of water at
     15oC, what is the final temperature of the system?
 Whenever we have a problem which deals with HEAT, we
 must recall the following relationship….
             HEATgained or lost = (Mass)(T)(Cp)
Keep in mind that the mass must be in grams, the temperature in
degrees Celsius, and the specific heat capacity in either cal/goC or
J/goC.
And since this problem involves TWO objects at DIFFERENT
TEMPERATURES being combined, we must also recall…...
                    HEATgained = HEATlost
 In other words, when we combine the two objects, the heat lost by
 the “warm” object will be gained by the “cool” object, or, put
 another way, the temperature of the “warm” object will decrease
 and the temperature of the “cool” object will increase.


                          <press any key to continue>
       If a piece of gold with a mass of 45.5 g and a
       temperature of 80.5oC is dropped into 192 g of water at
       15oC, what is the final temperature of the system?
               HEATgained or lost = (Mass)(T)(Cp)
                     HEATgained = HEATlost
 The problem asks us to determine the final temperature, so we will
 use “x” to represent that temperature.
 Looking at our list of COMMON SPECIFIC HEATS, we see
 that the Cp for gold = 0.13 J/goC and Cp for water = 4.18 J/goC.
 We now can re-write HEATgained = HEATlost ……...
    (Masswater)(Twater)(Cp water) = (Massgold)(Tgold)(Cp gold)
 Now we fill in this equation with our data……...
(192 g)(x - 15o C)(4.18 J/goC) = (45.5 g)(80.5o C - x)(0.13 J/goC)
Lastly, we solve for x…...
(802.56x - 12038.4) = (476.16 - 5.92x)
 808.48x = 12514.56
 x = 15.48o C
 The final temperature of the system will be 15.48o C.
                            <press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?


     Before continuing on with this tutorial, you should
     copy down the above problem and do the best you can
     do to solve it.
     Once you have the calculation completed, you should
     then go through the tutorial and see how you
     performed, paying particular attention to any areas
     where you may have made a mistake.




                           <press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?
   Whenever we have a problem which deals with HEAT, we
   must recall the following relationship….
              HEATgained or lost = (Mass)(T)(Cp)
  Keep in mind that the mass must be in grams, the temperature in
  degrees Celsius, and the specific heat capacity in either cal/goC or
  J/goC.
  And since this problem involves TWO objects at DIFFERENT
  TEMPERATURES being combined, we must also recall…...
                      HEATgained = HEATlost
   In other words, when we combine the two objects, the heat lost by
   the “warm” object will be gained by the “cool” object, or, put
   another way, the temperature of the “warm” object will decrease
   and the temperature of the “cool” object will increase.


                             <press any key to continue>
A piece of unknown metal with mass 14.9 g is heated to 100o C and
dropped into 75 g of water at 20o C. The final temperature of the
system is 28.5o C. What is the specific heat of the metal?
               HEATgained or lost = (Mass)(T)(Cp)
                    HEATgained = HEATlost
 The problem asks us to determine the specific heat of the metal, so
 we will use “x” to represent that specific heat.
 Looking at our list of COMMON SPECIFIC HEATS, we see
 that the Cp for water = 4.18 J/goC.
 We now can re-write HEATgained = HEATlost ……...
  (Masswater)(Twater)(Cp water) = (Massmetal)(Tmetal)(Cp metal)
 Now we fill in this equation with our data……...
(75 g)(28.5o C - 20o C)(4.18 J/goC) = (14.9 g)(100o C - 28.5o C)(x J/goC)
Lastly, we solve for x…...
(2664.75) = (1065.35x)
 x = 2.5 J/goC
 The specific heat of the metal is 2.5 J/goC.

				
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posted:10/4/2012
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