Measuring the Quantity of Heat

Measuring the Quantity of Heat

The specific heat capacity of solid aluminum (0.904 J/g/°C) is different than the specific heat capacity of solid iron (0.449 J/g/°C). This means that it would require more heat to increase the temperature of a given mass of aluminum by 1°C compared to the amount of heat required to increase the temperature of the same mass of iron by 1°C. In fact, it would take about twice as much heat to increase the temperature of a sample of aluminum a given amount compared to the same temperature change of the same amount of iron. This is because the specific heat capacity of aluminum is nearly twice the value of iron.

Heat capacities are listed on a per gram or per kilogram basis. Occasionally, the value is listed on a per mole basis, in which case it is called the molar heat capacity. The fact that they are listed on a per amount basis is an indication that the quantity of heat required to raise the temperature of a substance depends on how much substance there is. Any person who has boiled a pot of water on a stove, undoubtedly know this truth. Water boils at 100°C at sea level and at slightly lowered temperatures at higher elevations. To bring a pot of water to a boil, its temperature must first be raised to 100°C. This temperature change is achieved by the absorption of heat from the stove burner. One quickly notices that it takes considerably more time to bring a full pot of water to a boil than to bring a half-full of water to a boil. This is because the full pot of water must absorb more heat to result in the same temperature change. In fact, it requires twice as much heat to cause the same temperature change in twice the mass of water.

Specific heat capacities are also listed on a per K or a per °C basis. The fact that the specific heat capacity is listed on a per degree basis is an indication that the quantity of heat required to raise a given mass of substance to a specific temperature depends upon the change in temperature required to reach that final temperature. In other words, it is not the final temperature that is of importance, it is the overall temperature change. It takes more heat to change the temperature of water from 20°C to 100°C (a change of 80°C) than to increase the temperature of the same amount of water from 60°C to 100°C (a change of 40°C). In fact, it requires twice as much heat to change the temperature of a given mass of water by 80°C compared to the change of 40°C. A person who wishes to bring water to a boil on a stovetop more quickly should begin with warm tap water instead of cold tap water.

This discussion of specific heat capacity deserves one final comment. The term specific heat capacity is somewhat of a misnomer. The term implies that substances may have the ability to contain a thing called heat. As has been previously discussed, heat is not something that is contained in an object. Heat is something that is transferred to or from an object. Objects contain energy in a variety of forms. When that energy is transferred to other objects of different temperatures, we refer to transferred energy as heat or thermal energy. While it's not likely to catch on, a more appropriate term would be specific energy capacity.

Relating the Quantity of Heat to the Temperature Change

Specific heat capacities provide a means of mathematically relating the amount of thermal energy gained (or lost) by a sample of any substance to the sample's mass and its resulting temperature change. The relationship between these four quantities is often expressed by the following equation.

Q = m•C•ΔT

where Q is the quantity of heat transferred to or from the object, m is the mass of the object, C is the specific heat capacity of the material the object is composed of, and ΔT is the resulting temperature change of the object. As in all situations in science, a delta (∆) value for any quantity is calculated by subtracting the initial value of the quantity from the final value of the quantity. In this case, ΔT is equal to T_final - T_initial. When using the above equation, the Q value can turn out to be either positive or negative. As always, a positive and a negative result from a calculation has physical significance. A positive Q value indicates that the object gained thermal energy from its surroundings; this would correspond to an increase in temperature and a positive ΔT value. A negative Q value indicates that the object released thermal energy to its surroundings; this would correspond to a decrease in temperature and a negative ΔT value.

Knowing any three of these four quantities allows an individual to calculate the fourth quantity. A common task in many physics classes involves solving problems associated with the relationships between these four quantities. As examples, consider the two problems below. The solution to each problem is worked out for you. Additional practice can be found in the Check Your Understanding section at the bottom of the page.
 

Like any problem in physics, the solution begins by identifying known quantities and relating them to the symbols used in the relevant equation. In this problem, we know the following:

m = 450 g
C = 4.18 J/g/°C
T_initial = 15°C
T_final = 85°C

We wish to determine the value of Q - the quantity of heat. To do so, we would use the equation Q = m•C•ΔT. The m and the C are known; the ΔT can be determined from the initial and final temperature.

T = T_final - T_initial = 85°C - 15°C = 70.°C

With three of the four quantities of the relevant equation known, we can substitute and solve for Q.

Q = m•C•ΔT = (450 g)•(4.18 J/g/°C)•(70.°C)
Q = 131670 J
Q = 1.3x10⁵ J = 130 kJ (rounded to two significant digits)

Compared to the previous problem, this is a much more difficult problem. In fact, this problem is like two problems in one. At the center of the problem-solving strategy is the recognition that the quantity of heat lost by the water (Q_water) equals the quantity of heat gained by the metal (Q_metal). Since the m, C and ΔT values of the water are known, the Q_water can be calculated. This Q_water value equals the Q_metal value. Once the Q_metal value is known, it can be used with the m and ΔT value of the metal to calculate the Q_metal. Use of this strategy leads to the following solution:

Part 1: Determine the Heat Lost by the Water

Given:

m = 50.0 g
C = 4.18 J/g/°C
T_initial = 88.6°C
T_final = 87.1°C
ΔT = -1.5°C (T_final - T_initial)

Solve for Q_water:

Q_water = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(-1.5°C)
Q_water = -313.5 J (unrounded)
(The - sign indicates that heat is lost by the water)

Part 2: Determine the value of C_metal

Given:

Q_metal = 313.5 J (use a + sign since the metal is gaining heat)
m = 12.9 g
T_initial = 26.5°C
T_final = 87.1°C
ΔT = (T_final - T_initial )

Solve for C_metal:

Rearrange Q_metal = m_metal•C_metal•ΔT_metal to obtain C_metal = Q_metal / (m_metal•ΔT_metal)

C_metal = Q_metal / (m_metal•ΔT_metal) = (313.5 J)/[(12.9 g)•(60.6°C)]
C_metal = 0.40103 J/g/°C
C_metal = 0.40 J/g/°C (rounded to two significant digits)

 

Heat and Changes of State

The discussion above and the accompanying equation (Q = m•C•∆T) relates the heat gained or lost by an object to the resulting temperature changes of that object. As we have learned, sometimes heat is gained or lost but there is no temperature change. This is the case when the substance is undergoing a state change. So now we must investigate the mathematics related to changes in state and the quantity of heat.

To begin the discussion, let's consider the various state changes that could be observed for a sample of matter. The table below lists several state changes and identifies the name commonly associated with each process.

In the case of melting, boiling and sublimation, energy would have to be added to the sample of matter in order to cause the change of state. Such state changes are referred to as being endothermic. Freezing, condensation and deposition are exothermic; energy is released by the sample of matter when these state changes occur. So one might notice that a sample of ice (solid water) undergoes melting when it is placed on or near a burner. Heat is transferred from the burner to the sample of ice; energy is gained by the ice causing the change of state. But how much energy would be required to cause such a change of state? Is there a mathematical formula that might help in determining the answer to this question? There most certainly is.

The amount of energy required to change the state of a sample of matter depends on three things. It depends upon what the substance is, on how much substance is undergoing the state change, and upon what state change that is occurring. For instance, it requires a different amount of energy to melt ice (solid water) compared to melting iron. And it requires a different amount of energy to melt ice (solid water) as it does to vaporize the same amount of liquid water. And finally, it requires a different amount of energy to melt 10.0 grams of ice compared to melting 100.0 grams of ice. The substance, the process and the amount of substance are the three variables that affect the amount of energy required to cause a specific change in state. Use the widget below to investigate the effect of the substance and the process upon the energy change. (Note that the Heat of Fusion is the energy change associated with the solid-liquid state change.)

The values for the specific heat of fusion and the specific heat of vaporization are reported on a per amount basis. For instance, the specific heat of fusion of water is 333 J/gram. It takes 333 J of energy to melt 1.0 gram of ice. It takes 10 times as much energy - 3330 J - to melt 10.0 grams of ice. Reasoning in this manner leads to the following formulae relating the quantity of heat to the mass of the substance and the heat of fusion and vaporization.

For melting and freezing: Q = m•ΔH_fusion
For vaporization and condensation: Q = m•ΔH_vaporization

where Q represents the quantity of energy gained or released during the process, m represents the mass of the sample, ΔH_fusion represents the specific heat of fusion (on a per gram basis) and ΔH_vaporization represents the specific heat of vaporization (on a per gram basis). Similar to the discussion regarding Q = m•C•ΔT, the values of Q can be either positive or negative. Values of Q are positive for the melting and vaporization process; this is consistent with the fact that the sample of matter must gain energy in order to melt or vaporize. Values of Q are negative for the freezing and condensation process; this is consistent with the fact that the sample of matter must lose energy in order to freeze or condense.
 

As an illustration of how these equations can be used, consider the following two example problems.

The equation relating the mass (48.2 grams), the heat of fusion (333 J/g), and the quantity of energy (Q) is Q = m•ΔH_fusion. Substitution of known values into the equation leads to the answer.

Q = m•ΔH_fusion = (48.2 g)•(333 J/g)
Q = 16050.6 J
Q = 1.61 x 10⁴ J = 16.1 kJ (rounded to three significant digits)

Example Problem 3 involves a rather straightforward, plug-and-chug type calculation. Now we will try Example Problem 4, which will require a significant deeper level of analysis.

In this problem, the ice is melting and the liquid water is cooling down. Energy is being transferred from the liquid to the solid. To melt the solid ice, 333 J of energy must be transferred for every gram of ice. This transfer of energy from the liquid water to the ice will cool the liquid down. But the liquid can only cool as low as 0°C - the freezing point of the water. At this temperature the liquid will begin to solidify (freeze) and the ice will not completely melt.

We know the following about the ice and the liquid water:

Given Info about Ice:

m = 50.0 g
ΔH_fusion = 333 J/g

Given Info about Liquid Water:

C = 4.18 J/g/°C
T_initial = 26.5°C
T_final = 0.0°C
ΔT = -26.5°C (T_final - T_initial )

The energy gained by the ice is equal to the energy lost from the water.

Q_ice = -Q_{liquid water}

The - sign indicates that the one object gains energy and the other object loses energy. We can calculate the left side of the above equation as follows:

Q_ice = m•ΔH_fusion = (50.0 g)•(333 J/g)
Q_ice = 16650 J

Now we can set the right side of the equation equal to m•C•ΔT and begin to substitute in known values of C and ΔT in order to solve for the mass of the liquid water. The solution is:

16650 J = -Q_{liquid water}
16650 J = -m_{liquid water}•C_{liquid water}•ΔT_{liquid water}
16650 J = -m_{liquid water}•(4.18 J/g/°C)•(-26.5°C)
16650 J = -m_{liquid water}•(-110.77 J/°C)
m_{liquid water} = -(16650 J)/(-110.77 J/°C)
m_{liquid water} = 150.311 g
m_{liquid water} = 1.50x10² g (rounded to three significant digits)

Heating and Cooling Curves Revisited

On the previous page of Lesson 2, the heating curve of water was discussed. The heating curve showed how the temperature of water increased over the course of time as a sample of water in its solid state (i.e., ice) was heated. We learned that the addition of heat to the sample of water could cause either changes in temperature or changes in state. At the melting point of water, the addition of heat causes a transformation of the water from the solid state to the liquid state. And at the boiling point of water, the addition of heat causes a transformation of the water from the liquid state to the gaseous state. These changes in state occurred without any changes in temperature. However, the addition of heat to a sample of water that is not at any phase change temperatures will result in a change in temperature.

Now we can approach the topic of heating curves on a more quantitative basis. The diagram below represents the heating curve of water. There are five labeled sections on the plotted lines.

The three diagonal sections represent the changes in temperature of the sample of water in the solid state (section 1), the liquid state (section 3), and the gaseous state (section 5). The two horizontal sections represent the changes in state of the water. In section 2, the sample of water is undergoing melting; the solid is changing to a liquid. In section 4, the sample of water is undergoing boiling; the liquid is changing to a gas. The quantity of heat transferred to the water in sections 1, 3, and 5 is related to the mass of the sample and the temperature change by the formula Q = m•C•ΔT. And the quantity of heat transferred to the water in sections 2 and 4 is related to the mass of the sample and the heat of fusion and vaporization by the formulae Q = m•ΔH_fusion (section 2) and Q = m•ΔH_vaporization (section 4). So now we will make an effort to calculate the quantity of heat required to change 50.0 grams of water from the solid state at -20.0°C to the gaseous state at 120.0°C. The calculation will require five steps - one step for each section of the above graph. While the specific heat capacity of a substance varies with temperature, we will use the following values of specific heat in our calculations:

Solid Water: C=2.00 J/g/°C
Liquid Water: C = 4.18 J/g/°C
Gaseous Water: C = 2.01 J/g/°C

Finally, we will use the previously reported values of ΔH_fusion (333 J/g) and ΔH_vaporization (2.23 kJ/g).

Section 1: Changing the temperature of solid water (ice) from -20.0°C to 0.0°C.

Use Q₁ = m•C•ΔT

where m = 50.0 g, C = 2.00 J/g/°C, T_initial = -200°C, andT_final = 0.0°C

Q₁ = m•C•ΔT = (50.0 g)•(2.00 J/g/°C)•(0.0°C - -20.0°C)
Q₁ = 2.00 x10³ J = 2.00 kJ

Section 2: Melting the Ice at 0.0°C.

Use Q₂ = m•ΔH_fusion

where m = 50.0 g and ΔH_fusion = 333 J/g

Q₂ = m•ΔH_fusion = (50.0 g)•(333 J/g)
Q₂ = 1.665 x10⁴ J = 16.65 kJ
Q₂ = 16.7 kJ (rounded to 3 significant digits)

Section 3: Changing the temperature of liquid water from 0.0°C to 100.0°C.

Use Q₃ = m•C•ΔT

where m = 50.0 g, C = 4.18 J/g/°C, T_initial = 0.0°C, and T_final = 100.0°C

Q₃ = m•C•ΔT = (50.0 g)•(4.18 J/g/°C)•(100.0°C - 0.0°C)
Q₃ = 2.09 x10⁴ J = 20.9 kJ

Section 4: Boiling the Water at 100.0°C.

Use Q₄ = m•ΔH_vaporization

where m = 50.0 g and ΔH_vaporization = 2.23 kJ/g

Q₄ = m•ΔH_vaporization = (50.0 g)•(2.23 kJ/g)
Q₄ = 111.5 kJ
Q₄ = 112 kJ (rounded to 3 significant digits)

Section 5: Changing the temperature of liquid water from 100.0°C to 120.0°C.

Use Q₅ = m•C•ΔT

where m = 50.0 g, C = 2.01 J/g/°C, T_initial = 100.0°C, and T_final = 120.0°C

Q₅ = m•C•ΔT = (50.0 g)•(2.01 J/g/°C)•(120.0°C - 100.0°C)
Q₅ = 2.01 x10³ J = 2.01 kJ

The total amount of heat required to change solid water (ice) at -20°C to gaseous water at 120°C is the sum of the Q values for each section of the graph. That is,

Q_total = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Summing these five Q values and rounding to the proper number of significant digits leads to a value of 154 kJ as the answer to the original question.

In the above example, there are several features of the solution that are worth reflecting on:

We've learned here on this page how to calculate the quantity of heat involved in any heating/cooling process and in any change of state process. This understanding will be critical as we proceed to the next page of Lesson 2 on the topic of calorimetry. Calorimetry is the science associated with determining the changes in energy of a system by measuring the heat exchanged with the surroundings.