A sample of steam with a mass of 0.553 g and at a temperature of 100 degrees C condenses into an insulated container holding 4.25 g of water at 4.0 degrees C. Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?
heat lost by condensing steam + heat lost by water(from steam)changing to final T + heat gained by cold water moving to final T = 0
[mass steam x (-heat vap)] + [(mass steam(H2O) x specific heat H2O x (Tfinal-Tinitial)] + [mass cool H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
To determine the final temperature of the mixture, we can use the principle of conservation of energy.
First, let's calculate the heat energy gained by the water when it absorbs the heat from the steam during condensation.
The heat energy gained (Q) by the water can be calculated using the formula:
Q = m * c * ΔT
Where:
m = mass of the water (4.25 g)
c = specific heat capacity of water (4.18 J/g°C)
ΔT = change in temperature
The change in temperature (ΔT) can be calculated by subtracting the initial temperature (4.0°C) of the water from the final temperature.
Let's assume the final temperature is T°C.
So, ΔT = T - 4.0
Now, let's calculate the heat energy lost by the steam during condensation. Since no heat is lost to the surroundings, it is equal to the heat energy gained by the water.
Q_loss = Q_gain
The heat energy lost by the steam (Q_loss) can be calculated using the formula:
Q_loss = m * Hv
Where:
m = mass of the steam (0.553 g)
Hv = molar heat of vaporization of water (40.7 kJ/mol)
Since we are given the mass of steam, we need to convert it to molar quantity to use the molar heat of vaporization.
To convert the mass of steam to moles, we can use the formula:
n = m / M
Where:
n = number of moles
m = mass of the steam
M = molar mass of water (18.015 g/mol)
Now that we have the number of moles of steam, we can calculate the heat energy lost.
Q_loss = n * Hv
Next, we set up the equation Q_loss = Q_gain to equate the heat energy lost and gained by the steam and water.
m * Hv = m * c * (T - 4.0)
By substituting the values and solving for T, we can find the final temperature of the mixture.
To find the final temperature of the mixture, we can use the principle of conservation of energy. The heat lost by the steam (during condensation) will be gained by the water, resulting in an increase in its temperature.
Let's start by calculating the heat lost by the steam during condensation. We can use the formula:
q = m * L
Where:
q - heat lost/gained
m - mass of substance
L - latent heat of vaporization
The latent heat of vaporization for water is approximately 2260 J/g.
q_steam = m_steam * L
q_steam = 0.553 g * 2260 J/g
q_steam = 1245.28 J
Now, considering that this heat is gained by the water, we can calculate the change in temperature for the water using the formula:
q_water = m_water * C * ΔT
Where:
q_water - heat gained by water
m_water - mass of water
C - specific heat capacity of water
ΔT - change in temperature
The specific heat capacity of water is approximately 4.18 J/g°C.
q_water = 4.25 g * 4.18 J/g°C * ΔT
1245.28 J = 4.25 g * 4.18 J/g°C * ΔT
Now, let's rearrange the equation to solve for ΔT:
ΔT = 1245.28 J / (4.25 g * 4.18 J/g°C)
ΔT = 69.58°C
Therefore, the change in temperature for the water is 69.58°C. To find the final temperature, we can add this change in temperature to the initial temperature of the water:
Final temperature = Initial temperature + ΔT
Final temperature = 4.0°C + 69.58°C
Final temperature ≈ 73.58°C
Therefore, the final temperature of the mixture will be approximately 73.58°C.