HEAT CAPACITY AND HEAT OF REACTION

10.5. HEAT CAPACITY AND HEAT OF REACTION

Heat is a reactant or product in most chemical reactions. Before we consider including heat in a balanced chemical equation, first we must learn how to measure heat. When heat is added to a system, in the absence of

a chemical reaction the system may warm up, or a change of phase may occur. In this section change of phase will not be considered. Temperature and heat are not the same. Temperature is a measure of the intensity of the heat in a system. Consider the following experiment: Hold a lit candle under a pot of water with 0.5 in. of water in the bottom. Hold an identical candle, also lit, under an identical pot full of water for the same length of time. To which sample of water is more heat added? Which sample of water gets hotter?

The same quantity of heat is added to each pot, since identical candles were used for the same lengths of time. However, the water in the pot with less water in it is heated to a higher temperature. The greater quantity of water would require more heat to get it to the same higher temperature.

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 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 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

C. To warm 2 g of water 1 ◦

“change in.”

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.

EXAMPLE 10.14. How much heat does it take to raise the temperature of 10.0 g of water 20.1 ◦ C? Ans.

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? 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. 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?

( m)(c)

60.0 g)[4.184 J/(g· ◦ C)]

Note that the problem is to find the final temperature; 3.76 ◦ C is the temperature change.

C = 25.8 t ◦ final t initial t= C

C + 3.76 ◦

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 to 41.4 ◦ 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?

229 g)[4.184 J/(g · C)]

= t + t= 14.7 final ◦ initial

C + (−0.970 ◦

C) = 13.7 ◦ C

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

Ans.

50.0 g C

= 1640 kJ

12.0 g C

1 mol C

EXAMPLE 10.20. What rise in temperature will occur if 24.5 kJ of heat is added to 175 g of a dilute aqueous solution of sodium chloride [c = 4.10 J/g· ◦ C)] (a) by heating with a bunsen burner and (b) by means of a chemical reaction?

Ans. (a and b) The source of the heat does not matter; the temperature rise will be the same in either case. Watch out for the units!

( m)(c)

( 175 g)[4.10 J/(g· ◦

C)]

EXAMPLE 10.21. Calculate the heat of reaction per mole of water formed if 0.0500 mol of HCl and 0.0500 mol of NaOH are added to 15.0 g of water, all at 18.0 ◦

C to 54.3 ◦ C. The specific heat of the solution is 4.10 J/(g· ◦ C).

C. The solution formed is heated from 18.0 ◦

Ans. The law of conservation of mass allows us to calculate the mass of the solution:

15.0 g + 1.82 g + 2.00 g = 18.8 g

◦ C)](36.3 ◦

C) = 2800 J

2.80 kJ/(0.0500 mol water formed) = 56.0 kJ/mol