7.2: Heat

Skills to Develop

• To relate heat transfer to temperature change.

Heat is a familiar manifestation of energy. When we touch a hot object, energy flows from the hot object into our fingers, and we perceive that incoming energy as the object being “hot.” Conversely, when we hold an ice cube in our palms, energy flows from our hand into the ice cube, and we perceive that loss of energy as “cold.” In both cases, the temperature of the object is different from the temperature of our hand, so we can conclude that differences in temperatures are the ultimate cause of heat transfer.

Suppose we consider the transfer of heat from the opposite perspective—namely, what happens to a system that gains or loses heat? Generally, the system’s temperature changes. (We will address a few exceptions later.) The greater the original temperature difference, the greater the transfer of heat, and the greater the ultimate temperature change. The relationship between the amount of heat transferred and the temperature change can be written as

\[\text{heat} \propto ΔT \label{Eq1}\]

where ∝ means “is proportional to” and ΔT is the change in temperature of the system. Any change in a variable is always defined as “the final value minus the initial value” of the variable, so ΔT is T_final − T_initial. In addition, the greater the mass of an object, the more heat is needed to change its temperature. We can include a variable representing mass (m) to the proportionality as follows:

\[\text{heat} \propto mΔT \label{Eq2}\]

To change this proportionality into an equality, we include a proportionality constant. The proportionality constant is called the specific heat and is commonly symbolized by \(c\):

\[\text{heat} = mcΔT \label{Eq3}\]

Every substance has a characteristic specific heat, which is reported in units of cal/g•°C or cal/g•K, depending on the units used to express ΔT. The specific heat of a substance is the amount of energy that must be transferred to or from 1 g of that substance to change its temperature by 1°. Table \(\PageIndex{1}\) lists the specific heats for various materials.

Table \(\PageIndex{1}\): Specific Heats of Selected Substances

Substance	c (cal/g•°C)
aluminum (Al)	0.215
aluminum oxide (Al2O3)	0.305
benzene (C6H6)	0.251
copper (Cu)	0.092
ethanol (C2H6O)	0.578
hexane (C6H14)	0.394
hydrogen (H2)	3.419
ice [H2O(s)]	0.492
iron (Fe)	0.108
iron(III) oxide (Fe2O3)	0.156
mercury (Hg)	0.033
oxygen (O2)	0.219
sodium chloride (NaCl)	0.207
steam [H2O(g)]	0.488
water [H2O(ℓ)]	1.00

The proportionality constant c is sometimes referred to as the specific heat capacity or (incorrectly) the heat capacity.

The direction of heat flow is not shown in heat = mcΔT. If energy goes into an object, the total energy of the object increases, and the values of heat ΔT are positive. If energy is coming out of an object, the total energy of the object decreases, and the values of heat and ΔT are negative.

Example \(\PageIndex{1}\)

What quantity of heat is transferred when a 150.0 g block of iron metal is heated from 25.0°C to 73.3°C? What is the direction of heat flow?

SOLUTION

We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the iron is 73.3°C and the initial temperature is 25.0°C, ΔT is as follows:

ΔT = T_final − T_initial = 73.3°C − 25.0°C = 48.3°C

The mass is given as 150.0 g, and Table 7.3 gives the specific heat of iron as 0.108 cal/g•°C. Substitute the known values into heat = mcΔT and solve for amount of heat:

\(\mathrm{heat=(150.0\: g)\left(0.108\: \dfrac{cal} {g\cdot {^\circ C}}\right)(48.3^\circ C) = 782\: cal}\)

Note how the gram and °C units cancel algebraically, leaving only the calorie unit, which is a unit of heat. Because the temperature of the iron increases, energy (as heat) must be flowing into the metal.

Example \(\PageIndex{2}\)

A 10.3 g sample of a reddish-brown metal gave off 71.7 cal of heat as its temperature decreased from 97.5°C to 22.0°C. What is the specific heat of the metal? Can you identify the metal from the data in Table \(\PageIndex{1}\)?

SOLUTION

The question gives us the heat, the final and initial temperatures, and the mass of the sample. The value of ΔT is as follows:

ΔT = T_final − T_initial = 22.0°C − 97.5°C = −75.5°C

If the sample gives off 71.7 cal, it loses energy (as heat), so the value of heat is written as a negative number, −71.7 cal. Substitute the known values into heat = mcΔT and solve for c:

−71.7 cal = (10.3 g)(c)(−75.5°C)

\(c \,\mathrm{=\dfrac{-71.7\: cal}{(10.3\: g)(-75.5^\circ C)}}\)

c = 0.0923 cal/g•°C

This value for specific heat is very close to that given for copper in Table \(\PageIndex{1}\).

Summary

Heat transfer is related to temperature change.

Concept Review Exercise

1. Describe the relationship between heat transfer and the temperature change of an object.

Answer

1. Heat is equal to the product of the mass, the change in temperature, and a proportionality constant called the specific heat.

Exercises

1. A pot of water is set on a hot burner of a stove. What is the direction of heat flow?

2. How much energy in calories is required to heat 150 g of liquid H₂O from 0°C to 100°C?

3. If 250.0 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum? (Hint: First find temperature change.)

4. A sample of copper absorbs 145 cal of energy, and its temperature rises from 37.8°C to 41.7°C. What is the mass of the copper?

5. If 1.00 g of each substance in Table 7.3 were to absorb 100 cal of heat, which substance would experience the largest temperature change?

6. Determine the heat capacity of a substance if 23.6 g of the substance loses 199 cal of heat when its temperature changes from 37.9°C to 20.9°C. (Hint: If it "loses" heat, what is the sign of heat, positive or negative? If the temperature is going down, what is the sign for temperature change, ΔT = T_f - T_i? The sign of heat capacity should always be positive.)

Answers

1. Heat flows from the flame of the burner into the pot of water. (i.e. The flame loses/emits/gives off heat and the water gains/absorbs/takes in heat.)
2. 15,000 cal
3. 49.0°C
4. 404 g
5. Mercury would experience the largest temperature change.
6. 0.496 cal/g•°C; C = heat / [m ΔT] = -199 cal / [(23.6 g)(-17.0 ⁰C)] = 0.496 cal/g•°C