Specific and
Latent Heats

----------------------------
Joseph F. Alward, PhD
Department of  Physics  
University of the Pacific

 
Heat Unit:
One "calorie" (cal) = 4.186 Joules

 

 

   Specific Heat Capacity

Specific heat capacity, c:  

The amount of heat per gram which
must be gained or lost to raise or
lower the temperature 1 C.

Penny, marble, aluminum are at the
same temperature.

Which one stays hot longer after
removal from the bath?

 

 

 

 

 

   Specific Heat Capacity
DT  = Q/mc
---------------------------------

Same heat is absorbed.

Iron's ability to store
heat is less than water's.

Iron's temperature
rises more than does
the water's.

 

 

 

  Specific Heat Capacity
 
 
Q = mc DT
Substance   Specific Heat (cal/g-C)
 Water        1.00
 Aluminum        0.215
 Iron        0.108
Gold        0.031 

 

 

 

 

 

  Specific Heat Capacity

Which one stays hot longer
after being removed?  

Assume equal masses and
heat losses.

DT = Q/mc
Substance   Specific Heat
 Marble        0.90
 Aluminum        0.215
 Copper        0.092

 

 

 

 

   Why Hot Apple Pies Burn

               ccrust / capp
le  < 1
Comparing equal masses m,
and assuming the same Q loss:

DTapple = Q/mcapple

DTcrust = Q/mccrust

DTapple / DTcrust   = ccrust / capple

Apple's temperature drop is
smaller than the crust's.

(The substance with the higher
specific heat capacity stays hot
longer.)

 

 

  

  Hot Water Bottle
cwater    =  1.0  cal/g-C

calcohol =  0.57 cal/g-C
-----------------------------------
Assume both would lose
the same heat energy Q
in the same time.
-----------------------------------

DT = Q/mc

Which would be better to
keep you warm, a
hot-water bottle or a
hot-alcohol bottle?

 

 

 

 

 

 

 The Three Phases of Matter

 

 

 

 

   Phase Transformations

 

 

 

 

  Latent Heat of Fusion                                            latent:  undisclosed, beneath the surface    
 Heat of fusion of H2O:  80 cal/gram  
Fusion = melting or freezing
 Q = mLf  

Question:  How much thermal energy in joules must be absorbed by 50 g of ice at 0 C to melt it?

Answer:  50 (80) (4.186) = 16,744 J.  


Question:  How much thermal energy will be released when
50 g of water freezes?

Answer:  16,744 J.

 

 

 

 

  Keeping the Greenhouse Warm 
Placing large vats of water
in greenhouses protects
fruit from freezing; the heat
liberated when the water
freezes warms the air.

 

 

 

  Keeping the Wolf Cool
Panting wolf rids its body of
540 calories of heat energy
with each gram of water vapor
it exhales.

 

 

 

 

 

 

   Heat of Fusion
Heat of fusion of gallium = 80 cal/gram

Melting point = 29.8 C (85.6 F)

 

 

 

 

 

 

 

   Heat of Vaporization
Heat of Vaporization:  

Water:  L = 540 cal/g
---------------------------------------

Question:  How much heat
must be added to 20 grams
of water at 100 C to convert
it to steam?  

Answer:  
20 g (540 cal/g) = 10,800 cal

 

 

 

 

 

 

 

 

 

   Heat Numbers for H20
 Ice to ice:                      0.50     cal/g-C
 Ice to water:                     80     cal/g  
 Water to water:              1.0     cal/g-C   
 Water to steam:           540      cal/g    
 Steam to steam:          0.48     cal/g-C

 

 

 

 

 

 

 Temperature versus Heat for Water

 500 grams of -30 C ice is converted to 500 grams of 300 C steam.  How
 many calories are needed?

 

 

 

  Conservation of Energy
Heat lost by hot object =
Heat gained by liquid and cup

Example:  100 gram aluminum can with
200 grams of water, all at 20 C.  Add
50 gram of copper at 130 C.

What is the final temperature?
-------------------------------------------------------

Heat gained by water and aluminum
equals heat lost by copper:

Solution below:

 

 

 

 

 

 

 

  Conservation of Energy--continued
   

Heat gained by water
        = (200) (1.000) (T - 20)   (units dropped)
        = 200 T - 4000               (positive)

Heat gained by aluminum
         = (100 ) (0.215) (T - 20)    
         = 21.5 T - 430 

Heat lost by copper  
         = (50 ) (0.092) (130 - T)    
         = 598 - 4.6 T

Heat gained = Heat lost
(200 T - 4000) + (21.5 T - 430)      

         = 598 - 4.6 T

226.1 T = 5028

          T = 22.24 C