A 50.0g piece of iron an 152°C is dropped into 20.0g H2O(l) at 89°C in an open, thermally insulated container. How much water would you expect to vaporize, assuming no water splashes out? The specific heats of iron and water are 0.45 and 4.21 respectively, and ÄHvap= 40.7 kJ mol^-1 H2O.

The iron will lose this much energy upon cooling to 100C:

.45 J*g^-1*K^-1 * 50g * 52K = 1170 J

The water will gain this much energy when heating to 100C
4.21 J*g^-1*K^-1 * 20g * 11K = 926.2 J

The difference in those energies will be used to vaporize water:
1170 J - 926.2 J = 243.8 J

To vaporize water, you need:
(40700 J/mol) / (18 g/mol) = 2261 J/g

243.8 J / (2261 J/g) = 0.1 g of water will vaporize

To find out how much water would vaporize, we need to calculate the heat transferred from the iron to the water.

Step 1: Calculate the heat transferred from the iron to the water using the specific heat formula:

q = m × c × ΔT

Where:
q is the heat transferred
m is the mass of the substance
c is the specific heat
ΔT is the change in temperature

For the iron, m = 50.0 g, c = 0.45 J/g°C, and ΔT = (152 - 89) = 63°C

q(iron) = 50.0 g × 0.45 J/g°C × 63°C
q(iron) = 1417.5 J

For the water, m = 20.0 g, c = 4.21 J/g°C, and ΔT = (89 - 0) = 89°C

q(water) = 20.0 g × 4.21 J/g°C × 89°C
q(water) = 7514.98 J

Step 2: The heat released by the iron is equal to the heat absorbed by the water, but considering that some water will vaporize, we also need to account for the heat of vaporization.

q(iron) = q(water) + q(vaporization)

Let's assume x grams of water vaporize.

q(vaporization) can be calculated using the enthalpy of vaporization formula:

q(vaporization) = m × ΔHvap

Where:
q(vaporization) is the heat absorbed/released during vaporization
m is the mass of the substance
ΔHvap is the enthalpy of vaporization

For water, m = x g and ΔHvap = 40.7 kJ/mol = 40.7 × 10^3 J/mol

So, q(vaporization) = x g × 40.7 × 10^3 J/mol

Step 3: Now we can substitute the values into the equation:

1417.5 J = 7514.98 J + x g × 40.7 × 10^3 J/mol

Rearrange the equation to solve for x:

x g × 40.7 × 10^3 J/mol = 1417.5 J - 7514.98 J
x g × 40.7 × 10^3 J/mol = -6097.48 J

Divide both sides by 40.7 × 10^3 J/mol:

x g = -6097.48 J / (40.7 × 10^3 J/mol)

x g ≈ -0.15 g

Since we cannot have negative mass, the negative sign indicates that no water would vaporize.

To determine the amount of water that will vaporize, we can use the concept of heat transfer and the equation q = m * c * ΔT, where q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.

First, we need to find the heat transferred from the iron to the water. This can be calculated using the equation:

q_iron = m_iron * c_iron * ΔT_iron

Where:
m_iron = 50.0 g (mass of iron)
c_iron = 0.45 (specific heat of iron)
ΔT_iron = T_final - T_initial = (100°C - 89°C) = 11°C (change in temperature of iron)

Substituting the values, we have:
q_iron = 50.0 g * 0.45 * 11°C
q_iron = 247.5 cal

Next, we need to find the heat absorbed by the water. This can be calculated using the equation:

q_water = m_water * c_water * ΔT_water

Where:
m_water = 20.0 g (mass of water)
c_water = 4.21 (specific heat of water)
ΔT_water = T_final - T_initial = (100°C - 89°C) = 11°C (change in temperature of water)

Substituting the values, we have:
q_water = 20.0 g * 4.21 * 11°C
q_water = 928.2 cal

Since the system is thermally insulated, the heat lost by the iron is equal to the heat gained by the water. Therefore:

q_iron = q_water
247.5 cal = 928.2 cal

Now, we need to convert the heat transferred (q_water) to joules in order to use it to calculate the amount of water vaporized.

1 cal = 4.184 J

So, q_water = 928.2 cal * 4.184 J/cal
q_water = 3872.2368 J

Next, we need to calculate the moles of water vaporized using the enthalpy of vaporization (ΔH_vap) of water:

1 mol of H2O = 18.016 g
ΔH_vap = 40.7 kJ/mol

First, we need to convert the mass of water to moles:

moles_of_water = mass_of_water / molar_mass_of_water
moles_of_water = 20.0 g / 18.016 g/mol
moles_of_water = 1.1109 mol

Now, we can calculate the amount of water vaporized using the equation:

q_water = n_water * ΔH_vap
3872.2368 J = n_water * 40.7 kJ/mol * (1000 J/1 kJ)
3872.2368 J = n_water * 40,700 J
n_water = 3872.2368 J / 40,700 J
n_water = 0.095 mol

Finally, we need to convert the moles of water vaporized back to grams:

mass_of_water_vaporized = n_water * molar_mass_of_water
mass_of_water_vaporized = 0.095 mol * 18.016 g/mol
mass_of_water_vaporized = 1.715 g

Therefore, we would expect approximately 1.715 grams of water to vaporize in this scenario, assuming no water splashes out.