To solve this problem, we need to analyze the heat exchanges that occur when 1 kg of ice at 0°C, 1 kg of water at 100°C, and 1 kg of steam at 100°C are mixed in an insulated container. The goal is to find out how much of each phase remains after thermal equilibrium is reached.
Understanding the Heat Transfers
In this scenario, we have three substances at different phases and temperatures. The key concepts involved are the latent heats of fusion and vaporization, which are the energies required to change the state of a substance without changing its temperature.
- Latent heat of fusion: 80 cal/g (for ice melting to water)
- Latent heat of vaporization: 540 cal/g (for steam condensing to water)
Calculating Heat Transfers
Let's denote the masses of the substances:
- Mass of ice (m_ice) = 1 kg = 1000 g
- Mass of water (m_water) = 1 kg = 1000 g
- Mass of steam (m_steam) = 1 kg = 1000 g
Next, we will calculate the heat required for each phase change:
1. Ice Melting to Water
The heat required to melt the ice can be calculated using the formula:
Q_ice = m_ice × L_f
Where L_f is the latent heat of fusion.
Q_ice = 1000 g × 80 cal/g = 80000 cal
2. Steam Condensing to Water
The heat released when steam condenses into water is given by:
Q_steam = m_steam × L_v
Where L_v is the latent heat of vaporization.
Q_steam = 1000 g × 540 cal/g = 540000 cal
Heat Balance in the System
Now, we need to consider how these heat exchanges will balance out. The steam will release heat as it condenses, and this heat can be used to melt the ice and warm the resulting water.
Let’s analyze the situation:
- Heat released by steam when it condenses: 540000 cal
- Heat required to melt the ice: 80000 cal
After the ice melts, we have:
- Heat remaining from steam after melting ice: 540000 cal - 80000 cal = 460000 cal
Heating the Melted Ice (Water)
The melted ice (now water) will need to be heated from 0°C to 100°C. The heat required for this can be calculated as:
Q_heat = m_water × c × ΔT
Where c is the specific heat capacity of water (1 cal/g°C) and ΔT is the temperature change (100°C - 0°C = 100°C).
Q_heat = 1000 g × 1 cal/g°C × 100°C = 100000 cal
Now, we need to see if the remaining heat from the steam is enough to heat the melted ice:
- Remaining heat from steam: 460000 cal
- Heat required to heat the melted ice: 100000 cal
Since 460000 cal > 100000 cal, the melted ice can be heated to 100°C, and there will still be heat left over:
- Heat remaining after heating the melted ice: 460000 cal - 100000 cal = 360000 cal
Final Phase Changes
Now, we can use the remaining heat to convert some of the water at 100°C into steam:
Let’s calculate how much water can be converted into steam:
Using the latent heat of vaporization:
m_steam_produced = Q_remaining / L_v
m_steam_produced = 360000 cal / 540 cal/g ≈ 666.67 g
Final Amounts of Each Phase
After all the heat exchanges, we can summarize the final amounts:
- Ice remaining: 0 g (all melted)
- Water remaining: 1000 g (initial water) + 1000 g (melted ice) - 666.67 g (converted to steam) = 1333.33 g
- Steam produced: 666.67 g
In conclusion, after equilibrium is established, we have approximately:
- 0 g of ice
- 1333.33 g of water
- 666.67 g of steam
