Problem 101 - Chemistry

Question

What quantity of energy does it take to convert 0.500 \(\mathrm{kg}\) ice at \(-20.0^{\circ} \mathrm{C}\) to steam at \(250.0^{\circ} \mathrm{C} ?\) Specific heat capacities: ice, \(2.03 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} ;\) liquid, \(4.18 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C} ;\) steam, 2.02 \(\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) \(\Delta H_{\mathrm{vap}}=40.7 \mathrm{kJ} / \mathrm{mol} ; \Delta H_{\mathrm{fus}}=6.02 \mathrm{kJ} / \mathrm{mol} .\)

Step-by-Step Solution

Answer

It takes approximately \(1.70 \times 10^6 \, \mathrm{J}\) of energy to convert 0.500 kg of ice at -20°C to steam at 250°C.

Step 1: Convert mass to grams

We are given the mass of ice in kg, but specific heat capacities and enthalpies are given in terms of g and mol. Let's first convert the mass of ice to grams: \(m = 0.500 \mathrm{kg} \times 1000 \frac{\mathrm{g}}{\mathrm{kg}} = 500.0 \mathrm{g}\)

Step 2: Heat the ice to 0°C

Using the specific heat capacity of ice, calculate the energy needed to heat up the ice from -20°C to 0°C: \(Q_1 = m \times C_{ice} \times \Delta T_1 = 500.0 \mathrm{g} \times 2.03 \frac{\mathrm{J}}{\mathrm{g} \cdot ^{\circ}\mathrm{C}} \times (0 - (-20))^{\circ}\mathrm{C} = 20300.0 \, \mathrm{J}\)

Step 3: Melt the ice to water

For this, we first need to convert the mass of ice to moles, then use the enthalpy of fusion: \(\mathrm{Molar \, mass \, of \, H_2O} = 18.02 \frac{\mathrm{g}}{\mathrm{mol}}\) \(n = \frac{500.0 \mathrm{g}}{18.02 \frac{\mathrm{g}}{\mathrm{mol}}} = 27.75 \, \mathrm{mol}\) \(Q_2 = n \times \Delta H_{fus} = 27.75 \, \mathrm{mol} \times 6.02 \frac{\mathrm{kJ}}{\mathrm{mol}} = 166.995 \, \mathrm{kJ}\)

Step 4: Heat the water to 100°C

Now, calculate the energy needed to heat up the liquid water from 0°C to 100°C: \(Q_3 = m \times C_{liquid} \times \Delta T_2 = 500.0 \mathrm{g} \times 4.18 \frac{\mathrm{J}}{\mathrm{g} \cdot ^{\circ}\mathrm{C}} \times (100 - 0)^{\circ}\mathrm{C} = 209000.0 \, \mathrm{J}\)

Step 5: Vaporize water to steam

Now, calculate the energy needed to vaporize the water using the enthalpy of vaporization: \(Q_4 = n \times \Delta H_{vap} = 27.75 \, \mathrm{mol} \times 40.7 \frac{\mathrm{kJ}}{\mathrm{mol}} = 1129.425 \, \mathrm{kJ}\)

Step 6: Heat the steam to 250°C

Finally, calculate the energy needed to heat up the steam from 100°C to 250°C: \(Q_5 = m \times C_{steam} \times \Delta T_3 = 500.0 \mathrm{g} \times 2.02 \frac{\mathrm{J}}{\mathrm{g} \cdot ^{\circ}\mathrm{C}} \times (250 - 100)^{\circ}\mathrm{C} = 151000.0 \, \mathrm{J}\)

Step 7: Calculate the total energy

Now, add up all the energy values calculated in steps 2, 3, 4, 5, and 6: \(Q_{total} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 20300.0 \, \mathrm{J} + 166995 \, \mathrm{J} + 209000.0 \, \mathrm{J} + 1129425 \, \mathrm{J} + 151000.0 \, \mathrm{J} = 1704720.0 \, \mathrm{J}\) Thus, it takes approximately \(1.70 \times 10^6 \, \mathrm{J}\) of energy to convert 0.500 kg of ice at -20°C to steam at 250°C.

Key Concepts

Heat Capacity

Heat capacity is an essential concept in thermochemistry. It refers to the amount of energy needed to raise the temperature of a substance by a certain amount, typically one degree Celsius. In our exercise, we deal with specific heat capacities, which are heat capacities per gram of substance. This is important because it allows us to calculate how much energy is required to change the temperature of a specific mass of a substance.

For example, the specific heat capacity of ice is 2.03 J/g°C. This means you need 2.03 joules of energy to raise the temperature of one gram of ice by one degree Celsius. The exercise involves heating ice from -20°C to 0°C, which requires the formula:

• Formula: \[ Q = m \times C \times \Delta T \]
• Where:
  • \( Q \) is the heat energy.
  • \( m \) is the mass.
  • \( C \) is the specific heat capacity.
  • \( \Delta T \) is the change in temperature.

Understanding these components helps you determine how heat is absorbed or released in a system.

Phase Change

A phase change refers to the transition of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. These transitions occur when the temperature and pressure conditions of the substance change.

In the exercise, the phase changes include:

• Solid ice to liquid water, which requires melting.
• Liquid water to gaseous steam, which involves boiling or vaporization.

During a phase change, the temperature of the substance remains constant until the entire substance has transformed. This is because the heat energy is used to change the state rather than change the temperature. Hence, when you calculate the energy needed for a phase change, you focus on the enthalpy values relevant to that transition.

Enthalpy of Fusion

The enthalpy of fusion is a specific enthalpy change that describes the amount of energy required to convert a solid into a liquid at its melting point, without changing the temperature. In other words, it is the heat needed to melt one mole of solid substance. In our exercise, the focus is on melting ice into water.

The process requires calculating the moles of the substance and applying the enthalpy of fusion:

• Given: \( \Delta H_{fus} = 6.02 \text{ kJ/mol} \)
• To find the energy needed, use:
  • \[ Q = n \times \Delta H_{fus} \]

Converting grams to moles is critical because the enthalpy of fusion is provided per mole. Thus, \( n = \frac{m}{M} \), where \( M \) is the molar mass of water. This relation is why mass conversion is a crucial first step in these calculations.

Enthalpy of Vaporization

The enthalpy of vaporization is the energy required to transform a mole of liquid into a gas at its boiling point, again without a change in temperature. In our exercise, you calculate the energy necessary to convert water into steam.

For vaporization, the calculation involves:

• Given: \( \Delta H_{vap} = 40.7 \text{ kJ/mol} \)
• To find the energy needed, follow:
  • \[ Q = n \times \Delta H_{vap} \]

As with the enthalpy of fusion, it's vital to convert the mass of the water into moles. This allows the use of the enthalpy of vaporization specific to moles. By understanding vaporization, you can appreciate the energy involved in phase changes from a liquid state to a gaseous state.