The Foundations of Chemistry

By the unit factor approach,

Amount of heat
(205 g) (70.2°C)
6.02 104 J or 60.2 kJ

All units except joules cancel. To cool 205 g of water from 91.4°C to 21.2°C, it would be nec-

essary to remove exactly the same amount of heat, 60.2 kJ.

You should now work Exercises 54 and 55.

When two objects at different temperatures are brought into contact, heat flows from

the hotter to the colder body (Figure 1-18); this continues until the two are at the same

temperature. We say that the two objects are then in thermal equilibrium.The tempera-

ture change that occurs for each object depends on the initial temperatures and the rela-

tive masses and specific heats of the two materials.

EXAMPLE 1-19 Specific Heat

A 385-gram chunk of iron is heated to 97.5°C. Then it is immersed in 247 grams of water

originally at 20.7°C. When thermal equilibrium has been reached, the water and iron are both

at 31.6°C. Calculate the specific heat of iron.

Plan

The amount of heat gained by the water as it is warmed from 20.7°C to 31.6°C is the same as

the amount of heat lost by the iron as it cools from 97.5°C to 31.6°C. We can equate these

two amounts of heat and solve for the unknown specific heat.

Solution

Temperature change of water
31.6°C20.7°C
10.9°C

Temperature change of iron
97.5°C31.6°C
65.9°C

4.18 J



1 g°C

In specific heat calculations, we use the
magnitudeof the temperature change
(i.e., a positive number), so we subtract
the lower temperature from the higher
one in both cases.

38 CHAPTER 1: The Foundations of Chemistry

Figure 1-18 A hot object, such as a heated piece of metal (a), is placed into cooler water.

Heat is transferred from the hotter metal bar to the cooler water until the two reach the

same temperature (b). We say that they are then at thermal equilibrium.

(a) (b)