Skip to main content
Chemistry LibreTexts

7.2: Heat

  • Page ID
    68045
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Learning Objectives

    • 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 TfinalTinitial. 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 = TfinalTinitial = 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.

    Exercise \(\PageIndex{1}\)

    What quantity of heat is transferred when a 295.5 g block of aluminum metal is cooled from 128.0°C to 22.5°C? What is the direction of heat flow?

    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 = TfinalTinitial = 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 \(\PageIndex{1}\).

    Exercise \(\PageIndex{2}\)

    A 10.7 g crystal of sodium chloride (NaCl) had an initial temperature of 37.0°C. What is the final temperature of the crystal if 147 cal of heat were supplied to it?

    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. Some uncooked macaroni is added to a pot of boiling water. What is the direction of heat flow?

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

    4. How much energy in calories is required to heat 125 g of Fe from 25°C to 150°C?

    5. If 250 cal of heat were added to 43.8 g of Al at 22.5°C, what is the final temperature of the aluminum?

    6. If 195 cal of heat were added to 33.2 g of Hg at 56.2°C, what is the final temperature of the mercury?

    7. 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?

    8. A large, single crystal of sodium chloride absorbs 98.0 cal of heat. If its temperature rises from 22.0°C to 29.7°C, what is the mass of the NaCl crystal?

    9. 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?

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

    11. Determine the heat capacity of a substance if 23.6 g of the substance gives off 199 cal of heat when its temperature changes from 37.9°C to 20.9°C.

    12. What is the heat capacity of gold if a 250 g sample needs 133 cal of energy to increase its temperature from 23.0°C to 40.1°C?

    Answers

    1. Heat flows into the pot of water.

    1. 15,000 cal

    1. 49.0°C

    1. 404 g

    1. Mercury would experience the largest temperature change.

    1. 0.496 cal/g•°C


    7.2: Heat is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?