A copper of mass 400g is heated until it's temperature becomes 250 degree celcius.it is dropped into water weighing 200g at 50 degree celcius.calculate the mass of water vapour produced.
To solve this problem, we need to consider the heat gained by the water and the heat lost by the copper. We can use the heat transfer equation:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass
c is the specific heat capacity
ΔT is the change in temperature.
First, let's calculate the heat lost by the copper:
Qcopper = mcopper * ccopper * ΔTcopper
Given:
mcopper = 400g
ccopper = 0.385 J/g°C (specific heat capacity of copper)
ΔTcopper = 250°C - Tinitial (since we don't have the initial temperature, let's assume it's at room temperature, around 25°C)
Qcopper = 400g * 0.385 J/g°C * (250°C - 25°C)
Qcopper = 400g * 0.385 J/g°C * 225°C
Qcopper = 34,200 J
Now, let's calculate the heat gained by the water:
Qwater = mwater * cwater * ΔTwater
Given:
mwater = 200g
cwater = 4.186 J/g°C (specific heat capacity of water)
ΔTwater = Tfinal - Tinitial
Since we're trying to calculate the mass of water vapor produced, we don't need the final temperature (Tfinal). Instead, we need to consider the heat gained by the water as it reaches the boiling point and then gets converted to steam.
The specific heat capacity of water during the phase change from 50°C to the boiling point (100°C) is called the latent heat of vaporization (Lv). Lv for water is approximately 2260 J/g.
Let's calculate the heat gained by the water up to the boiling point:
Qwater = mwater * cwater * ΔTwater + mwater * Lv
Qwater = 200g * 4.186 J/g°C * (100°C - 50°C) + 200g * 2260 J/g
Qwater = 200g * 4.186 J/g°C * 50°C + 200g * 2260 J/g
Qwater = 20,930 J + 452,000 J
Qwater = 472,930 J
Since energy is conserved in this process, the heat lost by the copper is equal to the heat gained by the water:
Qcopper = Qwater
34,200 J = 472,930 J
34,200 J - 34,200 J = 472,930 J - 34,200 J
0 = 438,730 J
Now, let's calculate the mass of water vapor produced:
mwater vapor = Qwater / Lv
mwater vapor = 438,730 J / 2260 J/g
mwater vapor ≈ 194.06 g
Therefore, the mass of water vapor produced is approximately 194.06 grams.
To calculate the mass of water vapor produced, we need to use the principle of conservation of energy. We can assume that no heat is lost to the surroundings.
First, let's calculate the energy gained by the copper:
Q1 = m1 * c1 * ΔT1
Where:
m1 = mass of copper = 400g
c1 = specific heat capacity of copper = 0.385 J/g°C (approximate value)
ΔT1 = change in temperature of copper = 250°C
Q1 = 400g * 0.385 J/g°C * (250°C - initial temperature of copper)
Since the initial temperature is not given, we assume it to be at room temperature (25°C).
Q1 = 400g * 0.385 J/g°C * (250°C - 25°C) = 35750 J (joules)
Next, let's calculate the energy lost by the water:
Q2 = m2 * c2 * ΔT2
Where:
m2 = mass of water = 200g
c2 = specific heat capacity of water = 4.18 J/g°C (approximate value)
ΔT2 = change in temperature of water = (final temperature of water - initial temperature of water)
Q2 = 200g * 4.18 J/g°C * (50°C - initial temperature of water)
We want to find the initial temperature of the water. To do this, we equate the energy gained by the copper (Q1) to the energy lost by the water (Q2):
Q1 = Q2
400g * 0.385 J/g°C * (250°C - 25°C) = 200g * 4.18 J/g°C * (50°C - initial temperature of water)
Simplifying this equation, we can find the initial temperature of the water.
Solving for the initial temperature of the water, we find:
Initial temperature of water = 80°C
Finally, we need to determine the mass of water vapor produced. We use the equation:
Q3 = mLv
Where:
m = mass of water vapor produced (what we want to find)
Lv = latent heat of vaporization of water = 2260 J/g
The energy gained by the water (Q2) is equal to the energy required to evaporate the water (Q3):
Q2 = Q3
200g * 4.18 J/g°C * (50°C - 80°C) = m * 2260 J/g
Solving for the mass of water vapor produced (m):
m = (200g * 4.18 J/g°C * (50°C - 80°C)) / 2260 J/g
m ≈ 2.34g
Therefore, the mass of water vapor produced is approximately 2.34g.