Specific Heat of Selected Substances

Specific Heat J / g oC
Water (liquid) 4.18
Water (gas) 1.87
Water (solid) 2.06
Ethanol, C2H5OH(l) 2.438
Methane, CH4(g) 2.200
Isooctane, C8H18(l) 2.093
Aluminum, Al(s) 0.897
Table salt, NaCl(s) 0.865
Graphite, C(s) 0.714
Iron, Fe(s) 0.449
Silver, Ag(s) 0.235
Mercury, Hg(l) 0.139
Tungsten, W 0.132

  Heating Curve of Water

Calorimetry 1 WS     Calorimetry 2 WS

                                                                                                Causes of Change Outline

                Example 1:  In going from ice at -34oC to steam at 138oC, a sample of water absorbs 1.41 x 105 J.  Find the mass of the sample.

 

This problem will involve 5 steps:

Use X for the mass of the sample in all steps of the problem. 

Recall:   DT  =  Tf - Ti

Step 1:  Warm the ice from -34 oC to 0oC.

                equation:  DH1 = (mass)(Cp)( DT)                          Cp  ice (s) =  2.06 J/g oC      

                substitute:  DH1 = (X g)(2.06 J/g oC)(0 oC - (-34 oC))

Step 2:  Melt the ice at 0oC to water at 0oC

                equation:  DH2 = (mass)(Cf)                                      Cf  water  =  333 J/g       

                substitute:  DH2 = (X g)(333 J/g)

Step 3:  Warm the water from 0oC to 100 oC

                equation:  DH3 = (mass)(Cp)( DT)                          Cp  water (l)  =  4.184 J/g oC      

                substitute:  DH3 = (X g)(4.184 J/g oC)(100 oC -  0 oC)

Step 4:  Vaporize the water (boil) it from a liquid at 100oC to steam at 100oC

                equation:  DH4 = (mass)(Cv)                                      Cv  water  =  2256 J/g       

                substitute:  DH4 = (X g)(2256 J/g)

Step 5:  Superheat the steam from 100oC to 138oC.

                equation:  DH5 = (mass)(Cp)( DT)                          Cp  water (g)  =  1.87 J/g oC      

                substitute:  DH5 = (X g)(1.87 J/g oC)(138 oC -  100 oC)

Finally, the total heat absorbed by the sample will be equal to the sum of the heat absorbed by each step in the process:

DH1   +   DH1   +   DH1   +   DH1   +   DH1     =     1.41 x 105 J

Substitute:  (X g)(2.06 J/g oC)(0 oC - (-34 oC))  +  (X g)(333 J/g)  +  (X g)(4.184 J/g oC)(100 oC -  0 oC)  +  (X g)(2256 J/g)  +  (X g)(1.87 J/g oC)(138 oC -  100 oC)

Solve for X:  X  =  44.7 g

 

  Example 2:  If 20 g of silver at 350oC are mixed with 200 g of water at 30oC, find the final temperature of the system.

This problem will involve several steps. 

        Use X for the final temperature of the system.   Recall:   DT  =  Tf - Ti

The silver loses heat (energy = (-)) while the water gains heat (energy = (+).

Step 1:  - DHsilver = DHwater                     (Energy 'lost' by silver is equal to heat 'gained' by water)

                equation:  - (mass)(Cp)( DT)    =   (mass)(Cp)( DT)              Cp  silver (s) =  0.235 J/g oC     

                                                                                                                        Cp  water (s) =  4.184 J/g oC

              substitute:  - (20 g)(0.235 J/g oC)(X oC - 350 oC)  = (200 g)(4.184 J/g oC)(X oC - 30 oC)

                solve:  - 4.7X  +  1645  =  836X - 25,080

                combine like terms:  26.725   =  840.7X

                                    X  =  31.8oC