In this problem, we explore the process of vaporizing water that starts at an initial temperature of 90 degrees Celsius. To understand how much water vaporizes when heat energy is added, we can utilize a temperature versus heat (T vs. q) graph for water. This graph illustrates the relationship between temperature changes and phase changes as heat is added.

Initially, the water is at 90 degrees Celsius, which is below the boiling point of 100 degrees Celsius. As heat energy (denoted as q) is added, the temperature of the water increases until it reaches the boiling point. At this point, instead of continuing to increase in temperature, the water begins to vaporize into steam. The amount of water that vaporizes is represented as m_v, which is distinct from the total mass of the water, m.

To calculate the mass of water that vaporizes, we need to consider two key processes: the temperature increase to the boiling point and the phase change from liquid to vapor. The total heat added can be expressed with the equation:

q = m_cΔT + m_vL_v

Here, m_c is the mass of the water, ΔT is the change in temperature, and L_v is the latent heat of vaporization. For this scenario, we know:

• Total heat added, q = 5.89 × 10⁵ J
• Total mass of water, m = 0.6 kg
• Specific heat of water, c = 4186 J/(kg·K)
• Change in temperature, ΔT = 10 K (from 90°C to 100°C)
• Latent heat of vaporization, L_v = 2.256 × 10⁶ J/kg

Substituting these values into the equation, we first calculate the heat required to raise the temperature to the boiling point:

q_temp = m_cΔT = 0.6 kg × 4186 J/(kg·K) × 10 K = 25116 J

Next, we can substitute this value back into the total heat equation:

5.89 × 10⁵ J = 25116 J + m_v × 2.256 × 10⁶ J/kg

Rearranging the equation to solve for m_v gives:

m_v × 2.256 × 10⁶ J/kg = 5.89 × 10⁵ J - 25116 J

Calculating the right side results in:

m_v × 2.256 × 10⁶ J/kg = 5.64 × 10⁵ J

Finally, dividing both sides by the latent heat of vaporization allows us to find the mass of water that vaporizes:

m_v = (5.64 × 10⁵ J) / (2.256 × 10⁶ J/kg) ≈ 0.25 kg

This result indicates that approximately 0.25 kilograms of water vaporizes, confirming that not all of the initial water mass has turned into vapor, as this amount is less than the total mass of 0.6 kg. This analysis highlights the importance of understanding both temperature changes and phase transitions in thermodynamic processes.