How much heat is necessary to change a 52.0 g sample of water at 33.0 C into steam at 110.0 C?

q1 = heat needed to raise T liquid H2O from 33.0 to 100 C.

q1 = mass H2O x specific heat liquid H2O x Tfinal-Tinitial)

q2 = heat needed to vaporize all of the water at 100 to steam at 100.
q2 = mass H2O x heat vapoization

q3 = heat needed to raise T of steam at 100 C to steam at 110 C.
q3 = mass steam x specific heat steam x (Tfinal-Tinitial)
Total q = q1 + q2 + q3.

To calculate the amount of heat needed to change a substance from one state to another, we can use the formula:

Q = m * ΔH

Where:
Q is the heat energy
m is the mass of the substance
ΔH is the specific heat capacity or enthalpy change

In the case of water, we need to consider the following steps:

1. Heating the water from 33.0 °C to its boiling point (100.0 °C).
2. Changing the water at its boiling point into steam at the same temperature (100.0 °C).
3. Heating the steam from 100.0 °C to 110.0 °C.

Now let's calculate the heat energy required for each step and add them up.

Step 1: Heating the water from 33.0 °C to its boiling point (100.0 °C)

Q1 = m * c * ΔT

Where:
c is the specific heat capacity of water
ΔT is the change in temperature

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

ΔT1 = 100.0 °C - 33.0 °C = 67.0 °C

Q1 = 52.0 g * 4.18 J/g°C * 67.0 °C = 138,620 J

Step 2: Changing the water at its boiling point into steam at the same temperature (100.0 °C)

Q2 = m * ΔHv

Where:
ΔHv is the heat of vaporization of water

The heat of vaporization of water is approximately 2260 J/g.

Q2 = 52.0 g * 2260 J/g = 117,520 J

Step 3: Heating the steam from 100.0 °C to 110.0 °C

Q3 = m * c * ΔT

We'll consider the specific heat capacity of steam, which is approximately 2.09 J/g°C.

ΔT3 = 110.0 °C - 100.0 °C = 10.0 °C

Q3 = 52.0 g * 2.09 J/g°C * 10.0 °C = 10,876 J

Finally, to find the total heat energy required:

Q total = Q1 + Q2 + Q3 = 138,620 J + 117,520 J + 10,876 J = 267,016 J

Therefore, the amount of heat necessary to change the 52.0 g sample of water at 33.0 °C into steam at 110.0 °C is 267,016 Joules.

To calculate the amount of heat required to change a sample of water from one temperature to another, you can use the following formula:

Q = m * c * ΔT

where:
Q is the amount of heat,
m is the mass of the substance (water in this case),
c is the specific heat capacity of the substance,
ΔT is the change in temperature.

In this case, we need to calculate the heat required to change the water from 33.0 °C to 110.0 °C.

First, let's calculate the heat required to heat the water from 33.0 °C to its boiling point, which is 100.0 °C (since water boils at this temperature at atmospheric pressure):

Q1 = m * c * ΔT1

where:
m = 52.0 g (mass of the water)
c = 4.18 J/g°C (specific heat capacity of water)
ΔT1 = (100.0 °C - 33.0 °C) = 67.0 °C

Q1 = 52.0 g * 4.18 J/g°C * 67.0 °C

Now, let's calculate the heat required to convert the water at its boiling point (100.0 °C) to steam at 110.0 °C:

Q2 = m * ΔH

where:
ΔH is the heat of vaporization for water.

The heat of vaporization for water at atmospheric pressure is approximately 2260 J/g.

Q2 = 52.0 g * 2260 J/g

Now, let's calculate the total amount of heat required:

Q_total = Q1 + Q2

So, you can substitute the calculated values into the equation to find the answer.