Page 159 - Theory and Problems of BEGINNING CHEMISTRY

148 STOICHIOMETRY [CHAP. 10

The specific heat capacity of a substance is defined as the quantity of heat required to heat exactly 1 g
of the substance 1 C. Specific heat capacity is often called specific heat. Lowercase c is used to represent
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specific heat. For example, the specific heat of water is 4.184 J/(g· C). This means that 4.184 J will warm 1 g
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of water 1 C. To warm 2 g of water 1 C requires twice as much energy, or 8.368 J. To warm 1 g of water 2 C
requires 8.368 J of energy also. In general, the heat required to effect a certain change in temperature in a certain
sample of a given material is calculated with the following equation, where the Greek letter delta ( ) means
“change in.”
Heat required = (mass)(specific heat)(change in temperature) = (m)(c)( t)

Heat capacities may be used as factors in factor-label method solutions to problems. Be aware that there are two
units in the denominator, mass and temperature change. Thus, to get energy, one must multiply the heat capacity
by both mass and temperature change.

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EXAMPLE 10.14. How much heat does it take to raise the temperature of 10.0 g of water 20.1 C?
4.184 J
Ans. Heat = (m)(c)( t) = (10.0g) (20.1 C) = 841 J
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g· C
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EXAMPLE 10.15. How much heat does it take to raise the temperature of 10.0 g of water from 10.0 C to 30.1 C?
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Ans. This is the same problem as Example 10.14. In that problem the temperature change was specified. In this example,
the initial and final temperatures are given, but the temperature change is the same 20.1 C. The answer is again 841 J.
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EXAMPLE 10.16. What is the final temperature after 945 J of heat is added to 60.0 g of water at 22.0 C?
heat 945 J
Ans. t = = = 3.76 C
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(m)(c) (60.0g)[4.184 J/(g· C)]
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Note that the problem is to find the final temperature; 3.76 Cisthe temperature change.
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t final = t initial + t = 22.0 C + 3.76 C = 25.8 C
EXAMPLE 10.17. What is the specific heat of a metal alloy if 412 J is required to heat 44.0 g of the metal from 19.5 C
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to 41.4 C?
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heat 412 J 0.428 J
Ans. c = = =
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(m)( t) (44.0g)(41.4 C − 19.5 C) g· C
EXAMPLE 10.18. What is the final temperature of 229 g of water initially at 14.7 C from which 929 J of heat is removed?
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heat −929 J
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Ans. t = = =−0.970 C
mC (229 g)[4.184 J/(g · C)]
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t final = t initial + t = 14.7 C + (−0.970 C) = 13.7 C
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We note two things about this example. First, since the heat was removed, the value used in the equation was negative.
Second, the final temperature is obviously lower than the initial temperature, since heat was removed.
As was mentioned earlier in this section, heat is a reactant or product in most chemical reactions. It is possible
for us to indicate the quantity of heat in the balanced equation and to treat it with the rules of stoichiometry that
we already know.
EXAMPLE 10.19. How much heat will be produced by burning 50.0 g of carbon to carbon dioxide?
C + O 2 −→ CO 2 + 393 kJ
1 mol C 393 kJ
Ans. 50.0gC = 1640 kJ
12.0gC 1 mol C
We can use specific heat calculations to measure heats of reaction.