Question

Substance A has the following properties. \(\begin{array}{ll}\mathrm{mp} \text { at } 1 \mathrm{~atm}: & -20 .{ }^{\circ} \mathrm{C}\end{array}\) bp at 1 atm: \(\quad 85^{\circ} \mathrm{C}\) \(\Delta H_{\text {fus }}: \quad 180 . \mathrm{J} / \mathrm{g}\) \(\Delta H_{\mathrm{vap}}: \quad 500 . \mathrm{J} / \mathrm{g}\) \(c_{\text {solid: }}\) \(1.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) \(c_{\text {liquid }}\) \(2.5 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) \(c_{\mathrm{gas}}: \quad 0.5 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) At 1 atm, a 25 -g sample of \(\mathrm{A}\) is heated from \(-40 .{ }^{\circ} \mathrm{C}\) to \(100 .{ }^{\circ} \mathrm{C}\) at a constant rate of \(450 . \mathrm{J} / \mathrm{min} .\) (a) How many minutes does it take to heat the sample to its melting point? (b) How many minutes does it take to melt the sample? (c) Perform any other necessary calculations, and draw a curve of temperature vs. time for the entire heating process.

Short answer

It takes 53.89 minutes to heat the sample from -40°C to 100°C. Temperature vs. time graph will show plateaus at phase changes.

Step-by-step solution

Calculate the time to heat from -40°C to -20°C

For the temperature range from -40°C to -20°C (solid state), use the formula for heat energy: \[ q = mc\triangle T \]where \( m = 25 \text{ g} \), \( c_{\text{solid}} = 1.0 \text{ J/g} \times {^\text{∘} \text{C}} \), \( \triangle T = -20^\text{∘} \text{C} - -40^\text{∘} \text{C} = 20^\text{∘} \text{C} \). Then, \[ q = 25 \times 1.0 \times 20 = 500 \text{ J} \]The time required is then determined by dividing the heat by the heating rate:\[ \text{Time} = \frac{500 \text{ J}}{450 \text{ J/min}} = 1.11 \text{ min} \]

Calculate the time to melt the sample

Use the formula for the heat required to melt the sample: \[ q = m \times \triangle H_{\text{fus}} \]where \( \triangle H_{\text{fus}} = 180 \text{ J/g} \). Therefore, \[ q = 25 \times 180 = 4500 \text{ J} \]The time taken to melt the sample is \[ \text{Time} = \frac{4500 \text{ J}}{450 \text{ J/min}} = 10 \text{ min} \]

Calculate the time to heat from -20°C to 85°C

For the temperature range from -20°C to 85°C (liquid state), use the formula:\[ q = mc\triangle T \]where \( c_{\text{liquid}} = 2.5 \text{ J/g} \times {^\text{∘} \text{C}} \) and\( \triangle T = 85^\text{∘} \text{C} - (-20^\text{∘} \text{C}) = 105^\text{∘} \text{C} \). Then, \[ q = 25 \times 2.5 \times 105 = 6562.5 \text{ J} \]The time required is \[ \text{Time} = \frac{6562.5 \text{ J}}{450 \text{ J/min}} = 14.58 \text{ min} \]

Calculate the time to vaporize the sample

Use the formula for the heat required to vaporize the sample: \[ q = m \times \triangle H_{\text{vap}} \]where \( \triangle H_{\text{vap}} = 500 \text{ J/g} \). Therefore, \[ q = 25 \times 500 = 12500 \text{ J} \]The time taken to vaporize the sample is\[ \text{Time} = \frac{12500 \text{ J}}{450 \text{ J/min}} = 27.78 \text{ min} \]

Calculate the time to heat from 85°C to 100°C

For the temperature range from 85°C to 100°C (gas state), use the formula:\[ q = mc\triangle T \]where \( c_{\text{gas}} = 0.5 \text{ J/g} \times {^\text{∘} \text{C}} \) and\( \triangle T = 100^\text{∘} \text{C} - 85^\text{∘} \text{C} = 15^\text{∘} \text{C} \). Then, \[ q = 25 \times 0.5 \times 15 = 187.5 \text{ J} \]The time required is \[ \text{Time} = \frac{187.5 \text{ J}}{450 \text{ J/min}} = 0.42 \text{ min} \]

Sum of all times

Sum all the calculated times from Step 1 through Step 5 to find the total time required to heat the sample from -40°C to 100°C:\[ 1.11 \text{ min} + 10 \text{ min} + 14.58 \text{ min} + 27.78 \text{ min} + 0.42 \text{ min} = 53.89 \text{ min} \]

Draw the temperature vs time curve

Using the total times calculated for each stage of heating, draw the temperature versus time curve:1. From -40°C to -20°C over 1.11 minutes.2. Stay at -20°C for 10 minutes (melting phase).3. From -20°C to 85°C over 14.58 minutes.4. Stay at 85°C for 27.78 minutes (vaporization phase).5. From 85°C to 100°C over 0.42 minutes.The curve should show consistent heating up, plateaus at the phase changes, and continued heating up after the plateaus.

Key concepts

Heat Transfer

In thermodynamics, heat transfer refers to the movement of thermal energy from one object or substance to another. This process can occur through three main methods: conduction, convection, and radiation. When heating a substance, the heat energy is crucial for changing its temperature and state. In the given exercise, the substance A undergoes several heating stages, and heat transfer is continuously happening. The formula for heat energy, \[ q = mc \Delta T \] is used to calculate the amount of heat needed to change the temperature of the substance. Here, \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. It is important to apply the correct specific heat capacity depending on the state of the substance, whether it is solid, liquid, or gas.

Phase Transitions

Phase transitions are the changes of state from one phase to another, such as solid to liquid (melting) or liquid to gas (vaporization). During a phase transition, the temperature remains constant while the substance undergoes the change of state. For example, in the exercise, substance A melts at -20°C and vaporizes at 85°C.

• Melting: The transition from solid to liquid. The heat provided is used to break the molecular bonds within the solid structure.
• Vaporization: The transition from liquid to gas. The energy supplied is used to overcome the intermolecular forces within the liquid.

The equations for calculating the heat during these transitions are:\[ q = m \Delta H_{\text{fus}} \] for melting, and \[ q = m \Delta H_{\text{vap}} \] for vaporization.

Specific Heat Capacity

Specific heat capacity is a property of a substance that describes how much heat energy is needed to change the temperature of a unit mass of the substance by one degree Celsius. The specific heat capacities of substance A in solid, liquid, and gas states are given in the exercise.

• Solid: 1.0 J/g·°C
• Liquid: 2.5 J/g·°C
• Gas: 0.5 J/g·°C

The formula to calculate the heat required for raising the temperature of the substance is:\[ q = mc \Delta T \] where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Knowing the specific heat capacity helps in accurately determining the amount of heat required at different stages of heating.

Enthalpy of Fusion

Enthalpy of fusion (\( \Delta H_{\text{fus}} \)) is the amount of heat energy required to change a substance from solid to liquid at its melting point. This property is crucial during the melting phase of the heating process. In the given exercise, the enthalpy of fusion for substance A is 180 J/g.

• Melting Point Temperature: -20°C
• The heat required to melt the sample can be calculated using:
\[ q = m \Delta H_{\text{fus}} \] In this case for a 25g sample:\[ q = 25 \times 180 = 4500 \text{ J} \]To find out the time required, the formula: \[ \text{Time} = \frac{q}{\text{Heating Rate}} \] can be used, where the heating rate is 450 J/min in the exercise.

Enthalpy of Vaporization

Enthalpy of vaporization (\( \Delta H_{\text{vap}} \)) is the amount of heat energy required to change a substance from liquid to gas at its boiling point. This property plays a key role during the vaporization phase. For substance A, \( \Delta H_{\text{vap}} \) is given as 500 J/g.

• Boiling Point Temperature: 85°C
• The heat required to vaporize the sample is calculated using:
\[ q = m \Delta H_{\text{vap}} \] For a 25g sample:\[ q = 25 \times 500 = 12500 \text{ J} \]The time taken to vaporize the sample is given by:\[ \text{Time} = \frac{q}{\text{Heating Rate}} \] where the heating rate is 450 J/min.