How much energy (heat) is required to convert 248 g of water from 0oC to 154oC? Assume that the water begins as a liquid, that the specific heat of water is 4.184 J/g.oC over the entire liquid range, that the specific heat of steam is 1.99 J/g.oC, and the heat of vaporization of water is 40.79 kJ/mol.

q1 = heat needed to raise T from zero C to 100 C for the liquid.

q1 = mass H2O x specific heat x (Tfinal-Tinitial).

q2 = heat needed to vaporize liquid water at 100 C to steam at 100 C.
q2 = mass H2O x heat vap

q3 = heat needed to raise steam from 100 C to 154 C.
q3 = mass steam x specific heat x (Tfinal-Tinitial)
Then sum q1 + q2 + q3.
Note: Be VERY careful with the units. The problem quotes different units for the specific heat. You need to use the same unit which means converting them to the same, whichever you wish. The choice may be dictated by the units you have for mass H2O; i.e., you have grams, you might need to convert that to moles if you choose to use kJ/mol for sp. h.

I don't get it

To determine the total energy required to convert 248 g of water from 0oC to 154oC, we need to calculate the energy required for the following steps:

1. Heating the water from 0oC to 100oC.
2. Vaporizing the water at 100oC to form steam.
3. Heating the steam from 100oC to 154oC.

Let's calculate the energy required for each step:

Step 1: Heating the water from 0oC to 100oC.
The specific heat of water is given as 4.184 J/g.oC.
The temperature change is 100 - 0 = 100oC.
The mass of water is 248 g.

Energy required = mass * specific heat * temperature change
Energy required = 248 g * 4.184 J/g.oC * 100oC
Energy required = 104,099.2 J

Step 2: Vaporizing the water at 100oC to form steam.
We need to calculate the number of moles of water vaporized.
Molar mass of water = 2(1.01 g/mol) + 16.00 g/mol = 18.02 g/mol
Moles of water = mass / molar mass
Moles of water = 248 g / 18.02 g/mol
Moles of water = 13.776 mol

Heat of vaporization = 40.79 kJ/mol
Heat of vaporization = 40.79 kJ/mol * 1000 J/kJ
Heat of vaporization = 40,790 J/mol

Energy required = moles of water * heat of vaporization
Energy required = 13.776 mol * 40,790 J/mol
Energy required = 561,047.44 J

Step 3: Heating the steam from 100oC to 154oC.
The specific heat of steam is given as 1.99 J/g.oC.
The temperature change is 154 - 100 = 54oC.
The mass of water is 248 g (same as before).

Energy required = mass * specific heat * temperature change
Energy required = 248 g * 1.99 J/g.oC * 54oC
Energy required = 26,919.12 J

Total energy required = Energy for Step 1 + Energy for Step 2 + Energy for Step 3
Total energy required = 104,099.2 J + 561,047.44 J + 26,919.12 J
Total energy required = 692,065.76 J

Therefore, approximately 692,066 J of energy (heat) is required to convert 248 g of water from 0oC to 154oC.

To calculate the amount of energy required to convert 248g of water from 0oC to 154oC, we need to consider the different phases of water and the energy required for each phase change.

1. Heating water from 0oC to 100oC:
The specific heat capacity of water in the liquid state is given as 4.184 J/g.oC. Therefore, to raise the temperature of 248g of water by 100oC, we can use the formula:

Q1 = m * c * ΔT
Q1 = 248g * 4.184 J/g.oC * (100oC - 0oC)
Q1 = 104,371.2 J

2. Phase change from liquid water at 100oC to steam at 100oC:
At this point, the water is undergoing a phase change from liquid to gas. The energy required for this phase change is known as the heat of vaporization. Given that the heat of vaporization of water is 40.79 kJ/mol, we can calculate the energy required per gram of water:

Q2 = m * ΔHvap
Q2 = 248g * (40.79 kJ/mol / 18.015 g/mol)
Q2 = 555,784.1 J

3. Heating steam from 100oC to 154oC:
Now, the water is in the gaseous state (steam). The specific heat capacity of steam is given as 1.99 J/g.oC. Therefore, to raise the temperature of 248g of steam by 54oC, we can use the formula:

Q3 = m * c * ΔT
Q3 = 248g * 1.99 J/g.oC * (154oC - 100oC)
Q3 = 32,489.6 J

To find the total energy required, we sum up the three quantities:

Total energy = Q1 + Q2 + Q3
Total energy = 104,371.2 J + 555,784.1 J + 32,489.6 J
Total energy = 692,644.9 J

Therefore, the amount of energy (heat) required to convert 248g of water from 0oC to 154oC is 692,644.9 J.