A sample of steam with a mass of 0.552 and at a temperature of 100 condenses into an insulated container holding 4.35 of water at 6.0. Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?
To determine the final temperature of the mixture, we can use the principle of conservation of energy, which states that the total energy of an isolated system remains constant.
First, let's calculate the energy of the steam using the formula:
Energy = mass × specific heat capacity × temperature
Since the steam is condensed into an insulated container, the energy gained by the water will be equal to the energy lost by the steam. Therefore, we can set up the equation:
Energy gained by water = Energy lost by steam
Now, let's find the energy gained by the water using the formula:
Energy gained by water = mass × specific heat capacity × change in temperature
Given:
Mass of water (m1) = 4.35 kg
Specific heat capacity of water (c1) = 4186 J/kg°C (assuming it is pure water)
Initial temperature of water (t1) = 6.0°C
Let's calculate the energy gained by the water:
Energy gained by water = (m1) × (c1) × (final temperature - initial temperature)
= (4.35 kg) × (4186 J/kg°C) × (final temperature - 6.0°C)
= 18173.1 J × (final temperature - 6.0°C)
Now, let's find the energy lost by the steam using the formula:
Energy lost by steam = mass × specific heat capacity × change in temperature
Given:
Mass of steam (m2) = 0.552 kg
Specific heat capacity of steam (c2) = 2010 J/kg°C (assuming it is at constant pressure)
Initial temperature of steam (t2) = 100°C
Let's calculate the energy lost by the steam:
Energy lost by steam = (m2) × (c2) × (final temperature - initial temperature)
= (0.552 kg) × (2010 J/kg°C) × (final temperature - 100°C)
= 1111.512 J × (final temperature - 100°C)
Now, equating the energy gained by the water and the energy lost by the steam, we get:
18173.1 J × (final temperature - 6.0°C) = 1111.512 J × (final temperature - 100°C)
Simplifying the equation:
18173.1 J × (final temperature - 6.0°C) = 1111.512 J × final temperature - 111151.2 J
18173.1 J × final temperature - 108438.6 J = 1111.512 J × final temperature - 111151.2 J
Grouping the terms with final temperature on one side and the constant terms on the other side:
(18173.1 J - 1111.512 J) × final temperature = 108438.6 J - 111151.2 J
(17061.588 J) × final temperature = -2712.6 J
Dividing both sides by (17061.588 J):
final temperature = -2712.6 J / 17061.588 J
final temperature ≈ -0.159 °C
Since we obtained a negative value for the final temperature, it implies an error or that there might be some heat lost to the surroundings. Please double-check your calculations and ensure all the given values are accurate.
To solve this problem, we can use the principle of conservation of energy. The heat lost by the steam during condensation will be equal to the heat gained by the water to reach a final temperature.
The equation to calculate the heat gained or lost is given by:
Q = m * c * ΔT
where:
Q is the heat gained or lost (in Joules),
m is the mass of the substance (in kg),
c is the specific heat capacity of the substance (in J/kg·°C), and
ΔT is the change in temperature (in °C).
First, let's determine the specific heat capacity of steam and water.
The specific heat capacity of steam (csteam) is approximately 2,080 J/kg·°C.
The specific heat capacity of water (cwater) is approximately 4,186 J/kg·°C.
Now let's calculate the heat lost by the steam during condensation.
Qlost = msteam * csteam * ΔTsteam
Given:
msteam = 0.552 kg (mass of steam)
csteam = 2,080 J/kg·°C (specific heat capacity of steam)
ΔTsteam = 100 °C (change in temperature of steam, from 100°C to final temperature)
Qlost = 0.552 kg * 2,080 J/kg·°C * 100 °C
Qlost = 114,816 J
Next, let's calculate the heat gained by the water to reach the final temperature.
Qgained = mwater * cwater * ΔTwater
Given:
mwater = 4.35 kg (mass of water)
cwater = 4,186 J/kg·°C (specific heat capacity of water)
ΔTwater = final temperature - 6.0 °C (change in temperature of water, from 6.0°C to final temperature)
Since the system is insulated and no heat is lost to the surroundings,
Qgained = Qlost
Therefore,
mwater * cwater * ΔTwater = Qlost
Let's solve for ΔTwater:
ΔTwater = Qlost / (mwater * cwater)
ΔTwater = 114,816 J / (4.35 kg * 4,186 J/kg·°C)
Calculating ΔTwater,
ΔTwater ≈ 6.1 °C
Finally, we can find the final temperature of the mixture by adding ΔTwater to the initial temperature of the water.
Final temperature = 6.0 °C + ΔTwater
Final temperature ≈ 6.0 °C + 6.1 °C
Final temperature ≈ 12.1 °C
Therefore, the final temperature of the mixture will be approximately 12.1°C.