In a container of negligible mass, 0.100 kg of steam at 110°C and atmospheric pressure is added to 1.00 kg of water at 20.0°C.
(a) If no heat is lost to the surroundings, what is the final temperature of the system?
(b) At the final temperature, how many kg of steam are there, and how many kg of water?
Use the following pieces of data: The specific heat of water is 4190 J/kg·K. The specific heat of steam at atmospheric pressure is 1930 J/kg·K. The heat of vaporization of the water/steam phase transition is 2256000 J/kg.
To solve this problem, we need to apply the principle of conservation of energy.
(a) First, let's find the amount of heat transferred from the steam to the water.
Step 1: Calculate the heat absorbed by the water to reach the final temperature.
Using the specific heat equation: Q = m * c * ΔT
where:
- Q is the heat absorbed (in Joules),
- m is the mass of the water (1.00 kg),
- c is the specific heat of water (4190 J/kg·K),
- ΔT is the change in temperature.
ΔT = final temperature - initial temperature
ΔT = Tf - Ti
Given that the initial temperature (Ti) of the water is 20.0°C, we need to convert it to Kelvin (K).
Ti = 20.0°C + 273.15 = 293.15 K
Since no heat is lost to the surroundings, the heat absorbed by the water is equal to the heat transferred by the steam.
Step 2: Calculate the heat transferred by the steam.
Q = m * c * ΔT
where:
- Q is the heat transferred (in Joules),
- m is the mass of the steam (0.100 kg),
- c is the specific heat of steam at atmospheric pressure (1930 J/kg·K).
We don't know the change in temperature (ΔT) yet, so let's assume the final temperature is T.
ΔT = T - 110°C = T - 110 + 273.15 = T + 163.15 K
Since the heat absorbed by the water is equal to the heat transferred by the steam:
m_water * c_water * ΔT_water = m_steam * c_steam * ΔT_steam
Substituting the known values:
1.00 kg * 4190 J/kg·K * (T - 293.15 K) = 0.100 kg * 1930 J/kg·K * (T + 163.15 K)
Simplifying the equation:
4190 J * (T - 293.15) = 1930 J * (T + 163.15)
Solving for T:
4190T - 1231293 + 710543.5 = 1930T + 314319.5
2260T = 1551613
T ≈ 687.46 K
Therefore, the final temperature of the system is approximately 687.46 K.
(b) To find the masses of steam and water at the final temperature, we'll use the latent heat equation:
Q = m_water * L + m_water * c * ΔT_water + m_steam * c_steam * ΔT_steam
where:
- Q is the total heat transferred (in Joules),
- m_water is the mass of water (1.00 kg),
- L is the heat of vaporization of water/steam phase transition (2256000 J/kg).
Since we know Q from the previous calculations, we can rearrange the equation to solve for the mass of steam (m_steam):
m_steam = (Q - m_water * L - m_water * c * ΔT_water) / (c_steam * ΔT_steam)
Substituting the known values:
m_steam = (Q - 1.00 kg * 2256000 J/kg - 1.00 kg * 4190 J/kg·K * (T - 293.15 K)) / (1930 J/kg·K * (T + 163.15 K))
Plugging in T = 687.46 K and Q from the previous calculations, we can find the mass of steam.
m_steam = (Q - 1.00 kg * 2256000 J/kg - 1.00 kg * 4190 J/kg·K * (687.46 K - 293.15 K)) / (1930 J/kg·K * (687.46 K + 163.15))
Solving for m_steam gives the mass of steam at the final temperature.
Substitute Q and T into the expression to find m_steam.
Then, m_water = 1.00 kg - m_steam to find the mass of water at the final temperature.