Question

Does the amount of heat absorbed as \(1 \mathrm{~kg}\) of saturated liquid water boils at \(100^{\circ} \mathrm{C}\) have to be equal to the amount of heat released as \(1 \mathrm{~kg}\) of saturated water vapor condenses at \(100^{\circ} \mathrm{C}\) ?

Short answer

Answer: Yes, the amount of heat absorbed as 1 kg of saturated liquid water boils at 100°C is equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100°C, but with opposite signs.

Step-by-step solution

Understand the concepts of boiling and condensation

Boiling and condensation are phase changes between the liquid and gaseous states of a substance. When a substance undergoes boiling, it requires heat input to break the bonds between its particles. The heat used during this process is called the heat of vaporization. On the other hand, condensation is when the gaseous state of a substance converts back to liquid, and heat is released during this process.

Analyze the heat of vaporization and heat of condensation

The heat of vaporization is the amount of heat required to convert a unit mass of a substance from a liquid to a gaseous state at a constant temperature and pressure. The heat of condensation is the amount of heat released when a unit mass of a substance changes from gas to liquid. The heat of condensation for a substance is usually equal in magnitude to the heat of vaporization but of opposite sign.

Calculate the heat absorbed during boiling

Given that 1 kg of saturated liquid water boils at 100°C, the heat absorbed during this process can be calculated using the formula: \(Q = m \times L_v\) where Q is the heat absorbed, m is the mass of the substance (1 kg in this case), and L_v is the heat of vaporization for water. The heat of vaporization for water at 100°C is approximately \(2260 \mathrm{~kJ/kg}\). So, the heat absorbed can be calculated as: \(Q = 1 \mathrm{~kg} \times 2260 \mathrm{~kJ/kg} = 2260 \mathrm{~kJ}\)

Calculate the heat released during condensation

Similarly, when 1 kg of saturated water vapor condenses at 100°C, the heat released can be calculated using the formula: \(Q = -m \times L_v\) where Q is the heat released, and the negative sign indicates that heat is being released. So, the heat released can be calculated as: \(Q = -1 \mathrm{~kg} \times 2260 \mathrm{~kJ/kg} = -2260 \mathrm{~kJ}\)

Compare the heat absorbed and released

In this case, you can see that the magnitude of the heat absorbed during boiling is equal to the magnitude of the heat released during condensation: \(2260 \mathrm{~kJ} = |-2260 \mathrm{~kJ}|\) Therefore, the amount of heat absorbed as 1 kg of saturated liquid water boils at 100°C is equal to the amount of heat released as 1 kg of saturated water vapor condenses at 100°C, but with opposite signs.