Find the mass of 0 degree C ice that 15 g of 100 degree Csteam will completely melt.
In cooling down to liquid at 0 C, the steam will lose
Q = 540 cal/g*15g + 1.0cal/gC*100C*15g
= 8100 + 1500 = 9600 calories.
That amount of heat, transferred to the ice, can melt
9600 cal/80 cal/g = 120 g of ice.
The 540 cal/g and 80 cal/g numbers that I used are the heats of vaporization and fusion, respectively.
To find the mass of the ice that will completely melt, we need to first calculate the amount of heat absorbed by the ice to reach its melting point, and then divide that amount by the heat required to melt the ice.
Step 1: Calculate the heat absorbed by the ice to reach its melting point
The specific heat capacity of ice is 2.09 J/g·°C.
The initial temperature of the ice, Ti, is 0°C, and the final temperature, Tf, is 0°C (melting point of ice).
Q = m * c * ΔT
Where:
Q is the heat absorbed
m is the mass of the ice
c is the specific heat capacity of ice (2.09 J/g·°C)
ΔT is the change in temperature (Tf - Ti)
Q = m * 2.09 * (0 - 0)
Q = 0
This means that the ice does not absorb any heat to reach its melting point, as its initial and final temperatures are both at 0°C.
Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice is determined by the latent heat of fusion.
The latent heat of fusion for ice is 333.5 J/g.
Q = m * L
Where:
Q is the heat required
m is the mass of the ice
L is the latent heat of fusion (333.5 J/g)
We know that Q is also equal to the heat absorbed in Step 1, which is 0:
0 = m * 333.5
m = 0 / 333.5
m = 0 grams
Therefore, the mass of the 0°C ice that 15g of 100°C steam will completely melt is 0 grams.
To find the mass of ice that will be melted by 15 g of steam, we need to consider the energy transfer that occurs during the phase change from steam to ice.
First, let's determine the energy required to completely melt the ice. This energy is known as the heat of fusion for ice, which is the amount of energy required to change a substance from a solid to a liquid at its melting point.
The heat of fusion for ice is 334 J/g. This means that it takes 334 Joules of energy to melt one gram of ice.
Next, we need to find the total energy released by the 15 g of steam as it cools and condenses into water at 100 degrees Celsius. The heat of vaporization for water is the amount of energy required to change a substance from a liquid to a gas at its boiling point, which is 2260 J/g.
However, since the steam is initially at 100 degrees Celsius and needs to cool down to 0 degrees Celsius to completely melt the ice, we also need to consider the energy released during the cooling process. The specific heat capacity of water is approximately 4.18 J/g°C, meaning it takes 4.18 Joules of energy to raise the temperature of one gram of water by one degree Celsius.
To calculate the energy released during the cooling process, we can use the equation:
Q = m * C * ΔT
where Q is the energy released, m is the mass of the steam, C is the specific heat capacity of water, and ΔT is the change in temperature.
Assuming the steam condenses and cools all the way down to 0 degrees Celsius, ΔT = 100 - 0 = 100°C.
Therefore, the energy released during the cooling process is:
Q = (15 g) * (4.18 J/g°C) * (100°C) = 6270 J
Now, let's calculate the amount of ice that can be melted using this energy.
The energy required to melt one gram of ice is 334 J/g, so we can calculate the mass of ice melted by dividing the total energy released by the heat of fusion:
Mass of ice melted = (6270 J) / (334 J/g) = 18.77 g
Therefore, 15 g of steam at 100 degrees Celsius can completely melt approximately 18.77 g of ice at 0 degrees Celsius.