Measuring the Quantity of Heat Pt. 2 Physics January 22, 2013 Coach Stephens
Specific Heat of Fusion • 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•ΔHfusion • For vaporization and condensation: Q = m•ΔHvaporization
Q = m•ΔHfusion & Q = m•ΔHvaporization • where Q represents the quantity of energy gained or released during the process, m represents the mass of the sample, ΔHfusion represents the specific heat of fusion (on a per gram basis) and ΔHvaporization 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.
Example Problem #3 Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g. • The equation relating the mass (48.2 grams), the heat of fusion (333 J/g), and the quantity of energy (Q) is Q = m•ΔHfusion. Substitution of known values into the equation leads to the answer. • Q = m•ΔHfusion = (48.2 g)•(333 J/g)Q = 16050.6 JQ = 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.
Example Problem #4 What is the minimum amount of liquid water at 26.5 degrees C that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g. • 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 Information Given Info about Ice: • m = 50.0 gΔHfusion = 333 J/g Given Info about Liquid Water: • C = 4.18 J/g/°C Tinitial = 26.5°C Tfinal = 0.0°CΔT = -26.5°C (Tfinal - Tinitial ) • The energy gained by the ice is equal to the energy lost from the water. • Qice = -Qliquid water
Left Side of the Equation • The - sign indicates that the one object gains energy and the other object loses energy. • We can calculate the left side of the equation as follows: • Qice = m•ΔHfusion = (50.0 g)•(333 J/g)Qice = 16650 J
Right Side of the Equation 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 = -Qliquid water16650 J = -mliquidwater•Cliquidwater•ΔTliquid water16650 J = -mliquid water•(4.18 J/g/°C)•(-26.5°C)16650 J = -mliquid water•(-110.77 J/°C)mliquid water = -(16650 J)/(-110.77 J/°C)mliquid water = 150.311 gmliquid water = 1.50x102 g
Heating & Cooling Curve 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 gas 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 will result in a change in temperature.
Heating Curve of Water • 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.
Heating Curve • 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.
Heating Curve • 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•ΔHfusion (section 2) and Q = m•ΔHvaporization (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 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/°CLiquid Water: C = 4.18 J/g/°CGaseous Water: C = 2.01 J/g/°C • Finally, we will use the previously reported values of ΔHfusion (333 J/g) and ΔHvaporization (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, Tinitial = -200°C, andTfinal = 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•ΔHfusionwhere m = 50.0 g and ΔHfusion = 333 J/g • Q2 = m•ΔHfusion = (50.0 g)•(333 J/g)Q2 = 1.665 x104 J = 16.65 kJQ2 = 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, Tinitial = 0.0°C, and Tfinal = 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•ΔHvaporizationwhere m = 50.0 g and ΔHvaporization = 2.23 kJ/g • Q4 = m•ΔHvaporization = (50.0 g)•(2.23 kJ/g)Q4 = 111.5 kJQ4 = 112 kJ (rounded to 3 significant digits)
Section 5 • Changing the temperature of liquid water from 0.0°C to 100.0°C. • Use Q5 = m•C•ΔTwhere m = 50.0 g, C = 2.01 J/g/°C, Tinitial = 100.0°C, and Tfinal = 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
Final Step • 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, • Qtotal = 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.
Reflection • In the above example, there are several features of the solution that are worth reflecting on: • First: The lengthy problem was divided into parts, with each part representing one of the five sections of the graph. Since there were five Q values being calculated, they were labeled as Q1, Q2, etc. This level of organization is required in a multi-step problem such as this one. • Second: Attention was given to the +/- sign on ΔT. The change in temperature (or of any quantity) is always calculated as the final value of the quantity minus the initial value of that quantity. • Third: Attention was given to units throughout the course of the problem. Units of Q will either be in Joule or kiloJoule depending on which quantities are being multiplied. Failure to pay attention to units is a common cause of failure in problems like these. • Fourth: Attention was given to significant digits throughout the course of the problem. While this should never become the major emphasis of any problem in physics, it is certainly a detail worth attending to.
Conclusion • 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.