How much ice can be melted at 0oC by the condensation of 1.0 kg of steam at 100.oC?

(ΔHfus of ice = 6.01 kJ/mol; ΔHvap of water = 40.7 kJ/mol)

To determine how much ice can be melted by the condensation of 1.0 kg of steam, we can follow these steps:

Step 1: Calculate the amount of heat released from the condensation of 1.0 kg of steam.
Using the formula:
q = m * ΔHvap
where q is the heat released, m is the mass of steam, and ΔHvap is the heat of vaporization of water.

q = 1.0 kg * 40.7 kJ/kg
q = 40.7 kJ

Step 2: Convert the amount of heat released to moles of water (since the given enthalpies are in kJ/mol).
We can use the molar mass of water (H2O) to convert the mass of steam to moles.
The molar mass of H2O is 18.02 g/mol.

moles of water = (mass of steam / molar mass of water)
moles of water = (1000 g / 18.02 g/mol)
moles of water = 55.5 mol

Step 3: Calculate the amount of heat required to melt the ice.
The amount of heat required to melt the ice can be calculated using the equation:
q = n * ΔHfus
where q is the heat required, n is the number of moles of ice, and ΔHfus is the heat of fusion of ice.

From step 2, we know we have 55.5 moles of water.
Therefore, the number of moles of ice is also 55.5 moles.

q = 55.5 mol * 6.01 kJ/mol
q = 333.5 kJ

Step 4: Convert the amount of heat required to melt the ice back to mass.
To convert the heat to mass, we can use the equation:
mass = (heat released / ΔHfus)

mass = 333.5 kJ / 6.01 kJ/mol
mass = 55.49 mol

The molar mass of ice (H2O) is 18.02 g/mol.
Therefore, the mass of ice is:
mass = 55.49 mol * 18.02 g/mol
mass = 998.05 g

Therefore, approximately 998.05 grams or 1.0 kg of ice can be melted by the condensation of 1.0 kg of steam at 100.0°C.

To find the amount of ice that can be melted by the condensation of steam, we need to calculate the amount of heat released by the steam and compare it to the heat required to melt the ice.

First, let's calculate the amount of heat released by 1.0 kg of steam at 100°C when it condenses to water at 100°C. We can use the enthalpy of vaporization (ΔHvap) of water for this calculation.

The molar mass of water (H2O) is 18.02 g/mol, so the number of moles in 1.0 kg of steam is:
Number of moles = (mass of steam) / (molar mass of water)
Number of moles = (1000 g) / (18.02 g/mol) = 55.49 mol

The heat released by condensing the steam to water at 100°C can be calculated as:
Heat released = (number of moles) x (ΔHvap)
Heat released = (55.49 mol) x (40.7 kJ/mol)

Now, let's calculate the amount of ice that can be melted using the heat obtained from the condensation of steam. We can use the enthalpy of fusion (ΔHfus) of ice for this calculation.

The molar mass of ice (H2O) is also 18.02 g/mol, so the number of moles in the melted ice is:
Number of moles = (mass of melted ice) / (molar mass of ice)

To calculate the mass of melted ice, we need to convert the heat released to joules since ΔHfus is given in kJ/mol.
Heat released (in joules) = (heat released) x (1000 J/kJ)

Now, we can calculate the mass of melted ice as:
Mass of melted ice = (heat released in joules) / (ΔHfus)

Using the given values:
ΔHfus of ice = 6.01 kJ/mol
ΔHvap of water = 40.7 kJ/mol

Substituting these values, we can solve for the mass of melted ice:
Mass of melted ice = [(55.49 mol) x (40.7 kJ/mol) x (1000 J/kJ)] / (6.01 kJ/mol)

Calculating this expression will give us the amount of ice that can be melted by the condensation of 1.0 kg of steam at 100°C.

do it in phases:

consensation of steam
cooling of water from100C to 0+C
find the heat.

then set that equal to massice*Hfusion

solve for mass of ice.