Calculate the change in heat when 11.5 g of water vapor (steam) at 100.0°C condenses to liquid water and then cools to 5.00 °C

heats:

changeing to water at 100C:
mass*Hv calculate that
cooling the water from 100 to 5C
mass*specificheat*95

add the two heats

65677

To calculate the change in heat, we need to consider two separate processes: the condensation of steam to liquid water and the subsequent cooling of the liquid water.

First, let's calculate the heat released during the condensation of steam to liquid water:

1. Calculate the heat released during condensation:
The heat released during condensation is given by the equation:
q = m * ΔHvap
where q is the heat released, m is the mass of water vapor, and ΔHvap is the heat of vaporization.

The heat of vaporization for water is approximately 40.7 kJ/mol.

To calculate the moles of water vapor (steam), we can use the molar mass of water:
Molar mass of water = 18.015 g/mol

Number of moles of water vapor = mass / molar mass
Number of moles = 11.5 g / 18.015 g/mol

2. Calculate the heat released during condensation:
q = moles * ΔHvap
q = (number of moles) * (ΔHvap)
q = (11.5 g / 18.015 g/mol) * (40.7 kJ/mol)

Next, we need to calculate the heat released during cooling from 100.0°C to 5.00°C:

3. Calculate the heat released during cooling:
The specific heat capacity of water is approximately 4.18 J/(g·°C).

To calculate the heat released during cooling, we can use the equation:
q = m * c * ΔT
where q is the heat released, m is the mass of water, c is the specific heat capacity, and ΔT is the change in temperature.

Since water is already in its liquid state, we can assume the specific heat capacity remains constant.

4. Calculate the heat released during cooling:
q = m * c * ΔT
q = (mass of water) * (specific heat capacity) * (temperature change)
q = (11.5 g) * (4.18 J/(g·°C)) * (100.0°C - 5.00°C)

Finally, we can find the total change in heat by adding the heat released during condensation and the heat released during cooling:

Total change in heat = heat released during condensation + heat released during cooling

Please provide the value of the mass of water, in grams, so we can calculate the final result.

To calculate the change in heat when water vapor condenses to liquid water and then cools, we'll need to break it down into two steps:

1. Calculate the heat released during condensation.
2. Calculate the heat released during cooling.

Step 1: Calculate the heat released during condensation.

When water vapor condenses to liquid water, it releases heat. The heat released during condensation can be calculated using the formula:

q = m * ΔHv

where:
q is the heat released (in joules),
m is the mass of water vapor (in grams),
ΔHv is the heat of vaporization (in J/g).

The heat of vaporization for water is approximately 40.7 J/g.

Let's plug in the values:

m = 11.5 g
ΔHv = 40.7 J/g

q = 11.5 g * 40.7 J/g

q ≈ 468.05 J

Therefore, the heat released during condensation is approximately 468.05 J.

Step 2: Calculate the heat released during cooling.

To calculate the heat released during cooling, we'll use the formula:

q = m * Cp * ΔT

where:
q is the heat released (in joules),
m is the mass of the liquid water (in grams),
Cp is the specific heat capacity of water (in J/g°C),
ΔT is the change in temperature (in °C).

The specific heat capacity of water is approximately 4.18 J/g°C.

For this step, we need to find the mass of liquid water. This can be found using the equation:

m = m' - m''

where:
m' is the initial mass of water vapor,
m'' is the mass of condensed water vapor.

m' = 11.5 g (given)
m'' = 11.5 g (since all the water vapor condenses into liquid water)

m = 11.5 g - 11.5 g
m = 0 g

Since we have 0 g of liquid water in the second step, there is no heat released during cooling.

Therefore, the change in heat when 11.5 g of water vapor condenses to liquid water and then cools to 5.00 °C is approximately 468.05 J.