Question

Consider the following data for an unknown substance \(\mathrm{X} :\) $$\begin{array}{l}{\Delta H_{\mathrm{vap}}=20.00 \mathrm{kJ} / \mathrm{mol}} \\\ {\Delta H_{\mathrm{fus}}=5.00 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ $$\begin{array}{l}{\text { Specific heat capacity of solid }=3.00 \mathrm{Jg} \cdot^{\circ} \mathrm{C}} \\ {\text { Specific heat capacity of liquid }=2.50 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}} \\ {\text { Boiling point }=75.0^{\circ} \mathrm{C}} \\ {\text { Melting point }=-15.0^{\circ} \mathrm{C}} \\ {\text { Molar mass }=100.0 \mathrm{g} / \mathrm{mol}}\end{array}$$ In the heating of substance \(\mathrm{X}\) , energy (heat) is added at a constant rate of 450.0 \(\mathrm{J} / \mathrm{min}\) . At this rate, how long will it take to heat 10.0 \(\mathrm{g}\) of \(\mathrm{X}\) from \(-35.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} ?\)

Short answer

It will take approximately 4.67 minutes to heat 10.0 g of substance X from -35.0°C to 25.0°C at the given heat transfer rate of 450 J/min.

Step-by-step solution

Calculate heat required for heating the solid X from -35°C to -15°C

To calculate the heat required for this step, we can use the specific heat capacity of the solid along with the change in temperature: $$q_{1} = m C_{s} \Delta T$$ Where: \(q_{1}\) = the heat required in this step \(m\) = mass of substance X (10 g) \(C_{s}\) = specific heat capacity of solid X (3.00 J/g·°C) \(\Delta T\) = change in temperature (-15°C - (-35°C)) Plugging in the values, we get: $$q_{1} = (10 \, \text{g}) (3.00 \, \frac{\text{J}}{\text{g} \cdot ^{\circ} \text{C}}) (20 ^{\circ} \text{C}) = 600 \, \text{J}$$

Calculate heat required for melting the solid X at -15°C

To calculate the heat required for melting, we can use the enthalpy of fusion given in the exercise: $$q_{2} = m \Delta H_{\mathrm{fus}}$$ Where: \(q_{2}\) = the heat required in this step \(\Delta H_{\mathrm{fus}}\) = enthalpy of fusion (5.00 kJ/mol) First, we need to convert the mass of X to moles: $$n_\mathrm{X} = \frac{m}{M}$$ Where: \(n_\mathrm{X}\) = moles of substance X \(M\) = molar mass of X (100.0 g/mol) $$n_\mathrm{X} = \frac{10.0 \, \text{g}}{100.0 \, \frac{\text{g}}{\text{mol}}} = 0.1 \, \text{mol}$$ Now, we can find q2: $$q_{2} = (0.1 \, \text{mol}) (5000 \, \frac{\text{J}}{\text{mol}}) = 500 \, \text{J}$$

Calculate heat required for heating the liquid X from -15°C to 25°C

To calculate the heat required for this step, we can use the specific heat capacity of the liquid along with the change in temperature: $$q_{3} = m C_{l} \Delta T$$ Where: \(q_{3}\) = the heat required in this step \(C_{l}\) = specific heat capacity of liquid X (2.50 J/g·°C) \(\Delta T\) = change in temperature (25°C - (-15°C)) Plugging in the values, we get: $$q_{3} = (10\, \text{g}) (2.50 \, \frac{\text{J}}{\text{g} \cdot ^{\circ} \text{C}}) (40 ^{\circ}\text{C}) = 1000 \, \text{J}$$

Calculate the total heat required

Now, we can add up all the heat required in each step: $$q_\text{total} = q_{1} + q_{2} + q_{3} = 600 \, \text{J} + 500 \, \text{J} + 1000 \, \text{J} = 2100 \, \text{J}$$

Calculate the time required

Finally, we can use the heat transfer rate given in the exercise to calculate the time required to heat the 10 g of substance X: $$t = \frac{q_\text{total}}{\text{rate}}$$ Where: \(t\) = time required \(\text{rate}\) = heat transfer rate (450 J/min) $$t = \frac{2100 \, \text{J}}{450 \, \frac{\text{J}}{\text{min}}} = 4.67 \, \text{min}$$ Therefore, it will take approximately 4.67 minutes to heat 10.0 g of substance X from -35.0°C to 25.0°C at the given heat transfer rate.

Key concepts

Enthalpy of fusion

The enthalpy of fusion is the amount of energy needed to change a substance from a solid to a liquid at its melting point without changing the temperature. This is a crucial concept in thermochemistry because it helps us understand how much energy is required to overcome the forces holding the solid structure together.
For the substance X in our exercise, the enthalpy of fusion is given as 5.00 kJ/mol. This means that to melt one mole of substance X, we need to supply 5,000 J of energy. The enthalpy of fusion enables us to calculate the amount of heat necessary for phase changes, which is distinct from simply raising the temperature.

• Key point: Enthalpy of fusion relates to solid-to-liquid phase change.
• Measure in energy per mole (e.g., kJ/mol).
• Does not change temperature during the process.

Specific heat capacity

Specific heat capacity is a physical property that indicates how much energy is required to change the temperature of a substance by 1°C per gram. It plays a vital role in calculating energy changes during temperature variations without a phase change. This property varies depending on the state of the substance, whether it is solid, liquid, or gas.
In the given exercise, we see different specific heat capacities for the solid (3.00 J/g·°C) and the liquid (2.50 J/g·°C) forms of the substance X. This difference shows how much energy each state of the substance needs to raise the temperature of one gram by 1°C. The specific heat capacity is essential for assessing how a substance will react to heating.

• Enables calculation of heat required for temperature change.
• Different states (solid, liquid) have different capacities.
• Unit: J/g·°C or J/mol·°C based on context.

Phase change

Phase change refers to the transition between different states of matter, such as solid to liquid, liquid to gas, etc. During a phase change, the temperature of the substance remains constant, even as energy is being added or removed.
This is because the energy is used to change the state rather than increase the temperature. The common types of phase changes include melting, freezing, boiling, condensation, sublimation, and deposition.
In our exercise, we require understanding the phase change from solid to liquid (melting) when substance X reaches its melting point at -15°C. A similar concept applies during the boiling process, although the example focuses on melting. Calculating energy changes during phase transitions is essential since it differs from warming or cooling within the same phase.

• Involves energy without temperature change.
• Examples: melting, boiling, losing energy (e.g., freezing).
• Phase change energy calculated separately from temperature change.