As stated earlier, when an object is heated, it can get warmer, it can undergo a phase change, and it can undergo a chemical change. We have discussed the energy changes associated with warming and cooling objects. We now turn to the second possible outcome: a phase change.
If you heat a sample of ice at -10 °C at 1 atm pressure, it becomes warmer. If you heat ice at 0 °C (and 1 atm), however, initially it does not become warmer. Instead, the ice melts while maintaining a constant temperature of 0 °C, the melting point of water.
When a solid such as ice is heated and melts, the heat energy is used to overcome the forces holding the water molecules together in the solid phase. The heat energy is therefore changed into chemical potential energy. This potential energy can be reconverted to thermal energy when the forces between water molecules form again, that is, when liquid water freezes to form ice (Interactive Figure 5.3.5).
The energy to effect a phase change has been measured for a number of substances. The enthalpy (heat) of fusion (ΔHfus) for a substance is the energy needed to melt 1 g or 1 mol of that substance. The enthalpy (heat) of vaporization (ΔHvap) is the energy needed to vaporize 1 g or 1 mol of that substance.
Given the following information for mercury, Hg (at 1 atm), calculate the amount of heat needed (at 1 atm) to vaporize a 30.0-g sample of liquid mercury at its normal boiling point of 357 °C.
|boiling point = 357 °C||ΔHvap(357 °C) = 59.3 kJ/mol|
|melting point = -38.9 °C||ΔHfus(-38.9 °C) = 2.33 kJ/mol|
|specific heat (liquid) = 0.139 J/g • °C|
You are asked to calculate the amount of heat needed to vaporize a sample of liquid mercury at its normal boiling point.
You are given the mass of the mercury, its boiling and melting points, its heat of vaporization and heat of fusion, and the specific heat of the liquid.
This is a constant-temperature process in which the liquid is vaporized at its normal boiling point. Use the heat of vaporization of mercury to calculate the amount of heat needed to vaporize the liquid.
Is your answer reasonable? The quantity of liquid mercury being vaporized is less than one mol, so it should require less energy to vaporize than the heat of vaporization of mercury, the amount of energy required to vaporize one mole of liquid.
We now have the tools needed to calculate both the amount of thermal energy required to warm an object in a particular physical state and the amount of thermal energy required to change the physical state of a substance. Interactive Figure 5.3.6 explores the energy involved in heating 1-g samples of benzene and water over a large temperature range.
When 1-g samples of the solids at -20 °C are heated, the following processes occur between the indicated points on the heating curves:
As shown in the following example, the amount of heat required to heat a substance can be calculated when specific heat capacities, enthalpy of fusion, and enthalpy of vaporization values are known.
The following information is given for ethanol at 1 atm:
|boiling point = 78.40 °C||ΔHvap (78.40 °C) = 837.0 J/g|
|melting point = -114.5 °C||ΔHfus(-114.5 °C) = 109.0 J/g|
|specific heat (gas) = 1.430 J/g • °C||specific heat (liquid) = 2.460 J/g • °C|
A 33.50-g sample of liquid ethanol is initially at 13.50 °C. If the sample is heated at constant pressure (P = 1 atm), calculate the amount of energy needed to raise the temperature of the sample to 94.50 °C.
You are asked to calculate the total energy required to heat a sample of liquid ethanol.
You are given the mass of the ethanol sample, its initial and final temperature, and physical data for ethanol in the liquid and vapor phases.
This is a three step process: (1) raise the temperature of the liquid to its boiling point, (2) vaporize the liquid at its boiling point, and (3) raise the temperature of the resulting gas to the final temperature.
Calculate the amount of energy required to heat the liquid from 13.50 °C to 78.40 °C.
q(1) = m × cliquid × ΔT = (33.50 g)(2.460 J/g • °C)(78.40 °C - 13.50 °C) = 5348 J
Calculate the amount of energy required to vaporize the liquid at its boiling point.
q(2) = m × ΔHvap = (33.50 g)(837.0 J/g) = 2.804 × 104 J
q(3) = m × cgas × ΔT = (33.50 g)(1.430 J/g • °C)(94.50 °C - 78.40 °C) = 771.3 J
The total amount of energy needed to heat the sample of ethanol is the sum of the energy required for the three steps.
Is your answer reasonable? Over the temperature range in this problem, ethanol is heated, converted from liquid to vapor phase and then heated in the vapor phase. All three changes involve a great deal of energy, especially the conversion of a liquid to a vapor. The total amount of energy required to raise the temperature of the ethanol is expected to be large.