How many grams of ice could be melted by the energy obtained as 18.0 grams of steam is condensed at 100.0 degrees celcius and cooled to 0.0 degrees celcius?
The amount of heat liberated will be
Q = M*(Hv + 100 Cp)
where M = Mass (18.0g),
Hv = the heat of vaporization per gm (540 Cal/g), and
Cp is the specific heat of liquid water (1.00 Cal/g deg.C)
Q = 18*(540 + 100) = 11,520 Calories
Divide this by the heat of fusion of ice, 80 Cal/g, to get the amount of ice that will melt. This assumes that the ice started out at C.
I get 144 g. Check my numbers and method.
To solve this problem, we need to consider two processes: the condensation of steam to liquid water, and the subsequent cooling of the liquid water to ice.
First, let's calculate the energy released during the condensation of steam. The formula relating heat energy (q) and mass (m) is:
q = m * ΔHv
Where:
q = heat energy (in joules)
m = mass (in grams)
ΔHv = heat of vaporization (in J/g)
The heat of vaporization for water is approximately 40.7 J/g.
q = 18.0 g * 40.7 J/g
q = 732.6 J
Next, let's calculate the energy required to cool the liquid water from 100.0 degrees Celsius to 0.0 degrees Celsius. The formula relating heat energy (q) and mass (m) is:
q = m * c * ΔT
Where:
q = heat energy (in joules)
m = mass (in grams)
c = specific heat capacity (in J/g°C)
ΔT = change in temperature (in °C)
The specific heat capacity of water is approximately 4.18 J/g°C.
q = m * c * ΔT
q = 18.0 g * 4.18 J/g°C * (100.0 - 0.0) °C
q = 7524.0 J
Now, let's calculate the number of grams of ice that can be melted. The energy released during the steam condensation is equal to the energy required to cool the water and melt the ice:
q released = q required
732.6 J = 7524.0 J + m * ΔHf
Where:
q released = energy released during steam condensation (in J)
q required = energy required to cool water and melt the ice (in J)
m = mass of ice (in grams)
ΔHf = heat of fusion for water (in J/g)
The heat of fusion for water is approximately 334 J/g.
732.6 J = 7524.0 J + m * 334 J/g
Rearranging the equation to solve for m:
m = (732.6 J - 7524.0 J) / 334 J/g
m = -6791.4 J / 334 J/g
m ≈ -20.33 g
Since mass cannot be negative, we can conclude that the given amount of energy is not sufficient to melt any ice. Therefore, no grams of ice could be melted with the given conditions.
To calculate the amount of ice melted by the energy obtained from condensing steam, we need to consider the heat gained by the ice. The heat gained by ice can be calculated using the formula:
Q = m * ΔH
Where:
Q is the heat gained in Joules (J)
m is the mass in grams (g)
ΔH is the specific heat capacity of ice, which is 334 J/g°C
First, let's calculate the heat gained by the steam:
Q1 = m1 * ΔH1
Where:
m1 = 18.0 grams (mass of steam)
ΔH1 = specific heat capacity of water, which is 4.18 J/g°C (from steam to water)
Q1 = 18.0 g * 4.18 J/g°C
Q1 = 75.24 J
Next, let's calculate the heat lost by the water:
Q2 = m2 * ΔH2
Where:
ΔH2 is the specific heat capacity of water, which is also 4.18 J/g°C (from water at 100°C to water at 0°C)
Q2 = m2 * ΔH2
To calculate m2, we need to use the heat equation, considering that the heat gained by the ice is equal to the heat lost by the water:
Q1 = Q2
75.24 J = m2 * 4.18 J/g°C
Simplifying the equation, we can solve for m2:
m2 = 75.24 J / 4.18 J/g°C
m2 ≈ 18 grams
Therefore, approximately 18.0 grams of ice could be melted by the energy obtained as 18.0 grams of steam is condensed and cooled from 100.0°C to 0.0°C.