A sample of steam with a mass of 0.530g and at a temperature of 100 ∘C condenses into an insulated container holding 4.40g of water at 5.0∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C). What is the final temperature of the mixture, assuming no heat is lost in the surroundings?
To solve this problem, we need to use the principle of conservation of energy. Since the system is insulated, we can assume that the total energy before and after the steam condenses will remain constant.
The energy change can be calculated using the equation:
\( Q_{steam} = Q_{water} \)
The energy gained by the water can be calculated using the equation:
\( Q_{water} = m_{water} \cdot C_{water} \cdot \Delta T_{water} \)
Where:
\( m_{water} \) = mass of water (4.40g)
\( C_{water} \) = specific heat capacity of water (4.18 J/g⋅∘C)
\( \Delta T_{water} \) = change in temperature of water (final temperature - initial temperature)
The energy lost by the steam can be calculated using the equation:
\( Q_{steam} = m_{steam} \cdot \Delta H_{vap} \)
Where:
\( m_{steam} \) = mass of steam (0.530g)
\( \Delta H_{vap} \) = enthalpy of vaporization for steam (40.7 kJ/mol)
To calculate the final temperature of the mixture, we can set the two equations equal to each other and solve for \( \Delta T_{water} \).
\( m_{water} \cdot C_{water} \cdot \Delta T_{water} = m_{steam} \cdot \Delta H_{vap} \)
Rearranging the equation:
\( \Delta T_{water} = \frac{{m_{steam} \cdot \Delta H_{vap}}}{{m_{water} \cdot C_{water}}} \)
Substituting the given values:
\( \Delta T_{water} = \frac{{0.530g \cdot (40.7 \, \text{kJ/mol})}}{{4.40g \cdot (4.18 \, \text{J/g⋅∘C})}} \)
Now we can calculate the value of \( \Delta T_{water} \):
\( \Delta T_{water} = \frac{{21.581 \, \text{kJ/g}}}}{{18.352 \, \text{J/g⋅∘C}}} \)
\( \Delta T_{water} \approx 11.759 \, \text{∘C} \)
Finally, to find the final temperature of the mixture, we add the change in temperature to the initial temperature of the water:
\( \text{Final temperature} = \text{Initial temperature} + \Delta T_{water} \)
\( \text{Final temperature} = 5.0∘C + 11.759∘C \)
\( \text{Final temperature} \approx 16.759∘C \)
Therefore, the final temperature of the mixture, assuming no heat is lost in the surroundings, is approximately 16.759∘C.