# 5.7: Quantifying Heat Transfers: Temperature Changes

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##### Learning Objectives
• Define specific heat.
• Apply an equation to quantify the heat transfer that is associated with changing the temperature of a substance.

The primary objective of this chapter is to quantify the amount of heat that is transferred during physical and chemical changes. As stated previously, heat must be applied in order to overcome the attractive forces between a substance's constituent particles, in order to transform that substance from a solid to a liquid, and then, subsequently, from a liquid to a gas. Heat must also be transferred to a substance in order to raise its temperature. Because these transformations cannot occur simultaneously with one another, the heat transfers that are associated with these physical changes must be calculated separately.

This section will present and apply the equation that can be used to quantify the heat transfer that is associated with changing the temperature of a substance.

## Equation and Variables

In order to quantify the heat transfer, q, that is associated with changing the temperature of a substance, the mass, m, and specific heat, c, of that substance must be multiplied by the overall change in temperature, ΔT, which can be calculated by subtracting the initial temperature, Ti, of the substance from its final temperature, Tf, as shown in the equation below. Each of these quantities can be measured using multiple units. However, in order to be incorporated into this equation, the heat transfer must be recorded in calories (cal) or joules (J), the mass must be reported in grams (g), and the final and initial temperatures must be expressed in degrees Celsius (°C).

q = mc(Tf − Ti)

The specific heat, c, of a substance is a physical property that quantifies the amount of heat that is required required to raise the temperature of 1 gram of that substance by 1 degree Celsius. This definition is nearly identical to the definition of a calorie, which is the amount of heat that is required to raise the temperature of 1 gram of water by 1 degree Celsius. While the value of the calorie, which is the original unit that was utilized to measure heat, was derived based on experiments that involved water, the equation that is shown above must be applicable to a wide variety of chemicals. Therefore, specific heat, which is not associated with any particular chemical, by definition, is utilized in this equation. Because the variables in this equation are related multiplicatively, the unit for specific heat must incorporate the heat, mass, and temperature units that are indicated in the previous paragraph, in order to achieve unit cancelation. As a result, specific heats are reported in either cal/g·°C or J/g·°C. These units consist of both a numerator and a denominator, as indicated by the "/", and the "·" denotes the multiplicative relationship between the mass and temperature units. The specific heats for several compounds and elements are shown below in Table $$\PageIndex{1}$$.

Table $$\PageIndex{1}$$: Selected Specific Heat Values
Substance Specific Heat (cal/g·°C) Specific Heat (J/g·°C)
Water $$\left( l \right)$$ 1.00 4.184
Water $$\left( s \right)$$ 0.485 2.03
Water $$\left( g \right)$$ 0.447 1.87
Ammonia $$\left( g \right)$$ 0.488 2.04
Ethanol $$\left( l \right)$$ 0.588 2.46
Sodium Chloride $$\left( s \right)$$ 0.207 0.864
Aluminum $$\left( s \right)$$ 0.214 0.897
Carbon, graphite $$\left( s \right)$$ 0.169 0.709
Copper $$\left( s \right)$$ 0.0920 0.385
Gold $$\left( s \right)$$ 0.0308 0.129
Iron $$\left( s \right)$$ 0.108 0.452
Lead $$\left( s \right)$$ 0.0308 0.129
Mercury $$\left( l \right)$$ 0.0334 0.140
Silver $$\left( s \right)$$ 0.0562 0.235
Titanium $$\left( s \right)$$ 0.125 0.523

## Indicator Phrases

Because the equation that is shown above is used to quantify the heat transfer that is associated with changing the temperature of a substance, the phrases "change the temperature of," "raise the temperature of," and "lower the temperature of" indicate that this equation should be applied to solve a problem. While the first of these phrases is generic, the second and third variations of this indicator specify the direction in which the temperature change and, therefore, the associated heat transfer, occurs. If heat is added to a substance, its temperature will increase, and the corresponding heat transfer, q, will have a positive value. In contrast, the removal of heat from a substance will cause its temperature to decrease, and the associated heat transfer, q, will be negative.

The phrases that are given above are the primary indicators that the "q = mc(Tf − Ti)" equation should be used to solve a problem. The inclusion of temperature data or a reference to specific heat in a problem serve as secondary indicators that this equation should be applied to solve a problem. Finally, the final temperature, Tf, can be distinguished from the initial temperature, Ti, using the words "to" and "from" or "at," respectively.

## Calculations

For example, calculate how many calories of heat are required to raise the temperature of a 0.150 kilogram block of iron from 25.0 degrees Celsius to 73.3 degrees Celsius. The specific heat of iron is given in Table $$\PageIndex{1}$$.

The phrase "raise the temperature of," as well as the inclusion of temperature data and the reference to specific heat, indicates that the "q = mc(Tf − Ti)" equation should be used to solve this problem. Before this equation can be applied, each numerical quantity that is given in the problem must be assigned to a variable. Finally, in order to be incorporated into this equation, the validity of the units that are associated with the given numerical values must be confirmed. As stated above, the heat transfer must be recorded in calories (cal) or joules (J), the mass must be provided in grams (g), the specific heat must be expressed in either cal/g·°C or J/g·°C, and the final and initial temperatures must be reported in degrees Celsius (°C).

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.

Numerical Quantity Variable Unit Validity
q q
0.150 kg m
0.108 cal/g·°C c
73.3 °C Tf
25.0 °C Ti

Because q is the only variable that cannot be assigned to a numerical value in the given problem, heat transfer is the unknown quantity that will be calculated upon solving the "q = mc(Tf − Ti)" equation. Furthermore, the problem specifies that the final answer must be expressed in calories. Therefore, while Table $$\PageIndex{1}$$ lists two values for the specific heat of iron, 0.108 cal/g·°C and 0.452 J/g·°C, the first value must be incorporated into the equation that is indicated above, because its associated unit, cal/g·°C, is consistent with the unit that is specified for the unknown quantity, q. Of the remaining variables, only mass, m, is not reported in an acceptable unit. Therefore, as shown below, this quantity must be converted to grams before it can be incorporated into the "q = mc(Tf − Ti)" equation.

$${\text {0.150}} {\cancel{\rm{kg} }} \times$$ $$\dfrac{1,000 \; \rm{g} }{\cancel{\rm{kg} }}$$ = $${\text {150}} \; \rm{g}$$ ≈ $${\text {150.}} \; \rm{g}$$

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity Variable Unit Validity
q q
150. g m
0.108 cal/g·°C c
73.3 °C Tf
25.0 °C Ti

The quantities that are shown in the table above can now be incorporated into the "q = mc(Tf − Ti)" equation. When solving for q, order of operations dictates that subtraction occurs first. During the subsequent multiplication step, the mass and temperature units, "g" and "°C," respectively, are canceled, because both of these units are present in a numerator and a denominator in the third equation that is shown below. The unit that remains after these cancelations is "cal," which, per the information in the given problem, is the unit in which the unknown quantity, heat transfer, q, must be expressed. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

$$\text{q = mc}$$$$\left({\rm{T_f}} − {\rm{T_i}} \right)$$

$$\text{q} =$$ ($${150. \; \rm{g}}$$)$$\left({0.108 \; \dfrac{\rm{cal}} {\rm{g} \cdot \rm{^\circ C}}}\right)$$($${73.3 \; \rm{^\circ C} − 25.0 \; \rm{^\circ C}}$$)

$$\text{q} =$$ ($${150. \; \cancel{\rm{g}}}$$)$$\left({0.108 \; \dfrac{\rm{cal}} {\cancel{\rm{g}} \cdot \bcancel{\rm{^\circ C}}}}\right)$$($${48.3 \; \bcancel{\rm{^\circ C}}}$$)

$$\text{q} =$$ $${782.46 \; \rm{cal}} ≈ {782 \; \rm{cal}}$$

##### Example $$\PageIndex{1}$$

300. joules of heat must be removed to lower the temperature of 10.3 grams of an unknown metal from 208 degrees Fahrenheit to 295.2 Kelvin. Calculate the specific heat of the substance and compare its value the entries in Table $$\PageIndex{1}$$ to identify the metal.

Solution

The phrase "lower the temperature of," as well as the inclusion of temperature data and the reference to specific heat, indicates that the "q = mc(Tf − Ti)" equation should be used to solve this problem. Before this equation can be applied, each numerical quantity that is given in the problem must be assigned to a variable. Finally, in order to be incorporated into this equation, the validity of the units that are associated with the given numerical values must be confirmed. As stated above, the heat transfer must be recorded in calories (cal) or joules (J), the mass must be provided in grams (g), the specific heat must be expressed in either cal/g·°C or J/g·°C, and the final and initial temperatures must be reported in degrees Celsius (°C).

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.

Numerical Quantity Variable Unit Validity
-300. J q
10.3 g m
c c
295.2 K Tf
208 °F Ti

Because c is the only variable that cannot be assigned to a numerical value in the given problem, specific heat is the unknown quantity that will be calculated upon solving the "q = mc(Tf − Ti)" equation. Since the problem does not specify whether the final answer should be expressed in cal/g·°C or J/g·°C, the given unit for heat, "joules," is acceptable for this problem. Additionally, because the problem indicates that heat is removed in order to lower the temperature of the unknown metal, the value for the corresponding heat transfer, q, that is recorded in the table that is shown above has a negative value. Of the remaining variables, neither temperature is reported in an acceptable unit. Therefore, as shown below, both the final temperature, Tf,

295.2 K = TC + 273.15
TC = 22.05 °C ≈ 22.1 °C

and the initial temperature, Ti,

208 °F = 1.8TC + 32
176 ° = 1.8TC
TC = 97.77777... °C ≈ 97.8 °C

must both be converted to degrees Celsius before they can be incorporated into the "q = mc(Tf − Ti)" equation.

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.

Numerical Quantity Variable Unit Validity
-300. J q
10.3 g m
c c
22.1 °C Tf
97.8 °C Ti

The quantities that are shown in the table above can now be incorporated into the "q = mc(Tf − Ti)" equation. When solving for c, order of operations dictates that subtraction occurs first. During the subsequent multiplication of the mass and temperature values, no unit cancelation occurs, because there is no denominator component to either of these quantities. Instead, the multiplicative relationship between their units, "g" and "°C," respectively, is indicated by writing a "·" symbol between them. Finally, to solve for c, the heat transfer value on the left side of the equal sign must be divided by the quantity on the right side of the equality, which results in a cancelation of both negative signs. Again, no unit cancelation occurs, because the unit in the numerator, "J" does not match either of the units, "g" or "°C," that are in the denominator of the resultant fraction. Therefore, the unit that results from this division is "J/g·°C," which is a valid unit for expressing the unknown quantity, specific heat, c. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

$$\text{q = mc}$$$$\left({\rm{T_f}} − {\rm{T_i}} \right)$$

$${-300. \; \rm{J}} =$$ ($${10.3 \; \rm{g}}$$)$$\left(\text{c}\right)$$($${22.1 \; \rm{^\circ C} − 97.8 \; \rm{^\circ C}}$$)

$${-300. \; \rm{J}} =$$ ($${10.3 \; \rm{g}}$$)$$\left(\text{c}\right)$$($${-75.7 \; \rm{^\circ C}}$$)

$${-300. \; \rm{J}} =$$ ($${-779.71 \; \rm{g} \cdot \rm{^\circ C}}$$)$$\left(\text{c}\right)$$

$$\text{c} =$$ $${0.384758... \; \dfrac{\rm{J}} {\rm{g} \cdot \rm{^\circ C}}} ≈ {0.385 \; \dfrac{\rm{J}} {\rm{g} \cdot \rm{^\circ C}}}$$

This value corresponds to the specific heat of copper that is shown in Table $$\PageIndex{1}$$.

##### Exercise $$\PageIndex{1}$$

402 joules of heat are required to raise the temperature of 45.1 grams of cadmium to 62.7 degrees Celsius. Calculate the initial temperature of the cadmium, which has a specific heat of 0.0554 cal/g·°C.

The phrase "raise the temperature of," as well as the inclusion of temperature data and the reference to specific heat, indicates that the "q = mc(Tf − Ti)" equation should be used to solve this problem. Before this equation can be applied, each numerical quantity that is given in the problem must be assigned to a variable. Finally, in order to be incorporated into this equation, the validity of the units that are associated with the given numerical values must be confirmed. As stated above, the heat transfer must be recorded in calories (cal) or joules (J), the mass must be provided in grams (g), the specific heat must be expressed in either cal/g·°C or J/g·°C, and the final and initial temperatures must be reported in degrees Celsius (°C).

The numerical values that are given in the problem, the variables to which these quantities are assigned, and an indication of the validity of their corresponding units are shown in the following table.
Numerical Quantity Variable Unit Validity
402 J q
45.1 g m
0.0554 cal/g·°C c
62.7 °C Tf
Ti Ti
Because Ti is the only variable that cannot be assigned to a numerical value in the given problem, the initial temperature is the unknown quantity that will be calculated upon solving the "q = mc(Tf − Ti)" equation. While the problem does not specify a unit for the final answer, temperatures that are determined using this equation must be expressed in degrees Celsius, as stated above. All of the remaining variables are reported in acceptable units. However, the units that are associated with the given values for heat transfer, q, and specific heat, c, are not consistent with one another and, therefore, will not cancel when incorporated into the "q = mc(Tf − Ti)" equation. In order to remedy this discrepancy, one of these units must be converted to match the other. While altering either unit is acceptable, modifying the unit for heat transfer, q, is more straight-forward and, therefore, is shown below.

$${\text {402}} {\cancel{\rm{J} }} \times$$ $$\dfrac{1 \; \rm{cal} }{4.184 \; \cancel{\rm{J} }}$$ = $${\text {96.0803...}} \; \rm{cal}$$ ≈ $${\text {96.1}} \; \rm{cal}$$

The updated numerical values that are summarized in the following table are all expressed in the appropriate units and, therefore, can be utilized to solve the given problem.
Numerical Quantity Variable Unit Validity
96.1 cal q
45.1 g m
0.0554 cal/g·°C c
62.7 °C Tf
Ti Ti
The quantities that are shown in the table above can now be incorporated into the "q = mc(Tf − Ti)" equation. When solving for Ti, order of operations dictates that the multiplication of the mass and specific heat values occurs first. During this step, the mass unit is canceled, because "g" is present in a numerator and a denominator in the second equation that is shown below. Subsequently, the heat transfer value on the left side of the equal sign must be divided by the calculated quantity on the right side of the equality. This division causes the cancelation of the heat unit, "cal," which appears in both the numerator and the denominator of the resultant fraction. Finally, to solve for Ti, the final temperature must be subtracted from the calculated quantity on the left side of the equal sign. The calculated value corresponds to the negative value of the initial temperature, Ti. Therefore, the resultant quantities on both sides of the equal sign must be divided, or multiplied, by −1. Applying the correct number of significant figures to the calculated quantity results in the final answer that is shown below.

$$\text{q = mc}$$$$\left({\rm{T_f}} − {\rm{T_i}} \right)$$

$${96.1 \; \rm{cal}} =$$ ($${45.1 \; \cancel{\rm{g}}}$$)$$\left({0.0554 \; \dfrac{\rm{cal}} {\cancel{\rm{g}} \cdot \rm{^\circ C}}}\right)$$($${62.7 \; \rm{^\circ C} − \rm{T_i}}$$)

$${96.1 \; \cancel{\rm{cal}}} =$$ $$\left({2.49854 \; \dfrac{\cancel{\rm{cal}}} {\rm{^\circ C}}}\right)$$($${62.7 \; \rm{^\circ C} − \rm{T_i}}$$)

$${38.46246... \; \rm{^\circ C}} =$$ $${62.7 \; \rm{^\circ C} − \rm{T_i}}$$

$${-24.2375... \; \rm{^\circ C}} =$$ $${− \rm{T_i}}$$

$$\rm{T_i} =$$ $${24.2375 \; \rm{^\circ C}} ≈ {24.2 \; \rm{^\circ C}}$$

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