If 10g of steam at 110 c is pumped into an insulated vessel containing 100g of water at 20 c, what will be the equilibrium temperature of the mixture? spheat H2O(l)=4.18J/g*c, spheat h2O(g)=2.03J/g*c, heat of vap H2O(l)=2.260kJ/g
loss of heat moving steam from 110 to 100 is
q1 = mass steam x specific heat steam x (Tfinal-Tinitial) . Tf = 100; Ti = 110
q2 = loss of heat on condensing steam at 100 to liquid water at 100.
q2 = mass steam x heat vaporization.
I would do this in two steps.
First step:
q1 + q2 = mass water x specific heat x (Tf-Ti) where mass water is 100 and Ti= 100 C.
Then take that solution, use it as new Ti and use
mass 10g H2O x specific heat x (Tf - 100) + mass 100 g H2O x specific heat x (Tf-new Ti) = 0
Solve for Tf. I think something like 77 C is the answer.
To find the equilibrium temperature of the mixture, we need to calculate the heat gained by the water and the heat lost by the steam until they reach the same temperature.
Step 1: Find the heat gained by the water
Q1 = mass_water * specific_heat_water * (final_temperature - initial_temperature)
Q1 = 100g * 4.18 J/g°C * (T - 20°C)
Step 2: Find the heat lost by the steam
Q2 = mass_steam * specific_heat_steam * (initial_temperature - final_temperature)
Q2 = 10g * 2.03 J/g°C * (110°C - T)
Step 3: Set Q1 equal to Q2 and solve for T
Q1 = Q2
100g * 4.18 J/g°C * (T - 20°C) = 10g * 2.03 J/g°C * (110°C - T)
Step 4: Convert heat of vaporization from kJ/g to J/g
heat_of_vaporization = 2.260 kJ/g * 1000 J/1 kJ = 2260 J/g
Step 5: Add the heat gained during vaporization to Q1
Q1 = 100g * 4.18 J/g°C * (T - 20°C) + 10g * 2260 J/g
Step 6: Set Q1 equal to Q2 and solve for T
100g * 4.18 J/g°C * (T - 20°C) + 10g * 2260 J/g = 10g * 2.03 J/g°C * (110°C - T)
Step 7: Simplify and solve for T
(T - 20) * 100 * 4.18 + 10 * 2260 = 10 * 2.03 * (110 - T)
Step 8: Solve the equation for T
Simplifying the equation will give us:
418T - 8360 + 22600 = 226 * (110 - T)
418T + 14240 = 22600 - 226T
Step 9: Combine like terms
418T + 226T = 22600 - 14240
644T = 8360
T ≈ 13.0°C
Therefore, the equilibrium temperature of the mixture is approximately 13.0°C.
To find the equilibrium temperature of the mixture, we need to consider the heat transfer that occurs when steam condenses and transfers its heat to the water.
First, let's calculate the heat transfer when steam condenses to liquid water. The heat of vaporization of water is given as 2.260 kJ/g. Since 10 g of steam is condensed, the heat transfer can be calculated as follows:
Heat transfer = mass of steam x heat of vaporization
= 10 g x 2.260 kJ/g
Next, let's calculate the heat transfer when the liquid water increases in temperature. The specific heat capacity of water (liquid) is given as 4.18 J/g*°C. We need to calculate the temperature change required for the water to reach equilibrium with the condensed steam.
To do this, we'll assume that the final temperature of the mixture is T. The heat transfer from 100 g of water at 20°C to reach T can be calculated as follows:
Heat transfer = mass of water x specific heat capacity x temperature change
= 100 g x 4.18 J/g*°C x (T - 20°C)
Since the system is insulated, the heat transfer from the steam to the water is equal to the heat transfer from the water to the steam. Therefore, we can set up an equation as follows:
10 g x 2.260 kJ/g = 100 g x 4.18 J/g*°C x (T - 20°C)
Now, we can solve this equation to find the equilibrium temperature T.
First, convert the heat of vaporization to joules:
2.260 kJ/g = 2.260 x 10^3 J/g
10 g x (2.260 x 10^3 J/g) = 100 g x 4.18 J/g*°C x (T - 20°C)
(2.260 x 10^4 J) = 100 x 4.18 J/g*°C x (T - 20°C)
Divide both sides by (100 x 4.18 J/g*°C):
(2.260 x 10^4 J) / (100 x 4.18 J/g*°C) = T - 20°C
Simplify the equation:
541.63°C = T - 20°C
Now solve for T:
T = 541.63°C + 20°C
T = 561.63°C
Therefore, the equilibrium temperature of the mixture will be approximately 561.63°C.