Help me solve this problem

Question

How much heat is required to warm 35.21g of liquid water at 88.53°C to steam (gaseous water) at 124.67°C?

Specific Heat (ice) = 37.6 J/K*mol; Specific Heat (water) = 75.3 J/K*mol; Specific Heat (steam) = 35.8 J/K*mol; .

ΔHfusion (ice) = 6.01 kJ/mol; ΔHvaporization (water) = 40.67 kJ/mol; ΔHsublimation (ice) = 46.7 kJ/mol.

Answer 1

Step 1) convert grams of H2O to moles

35.21g HsO x 1mole/18g = 1.96 moles of H20

This requires calculations for 2 heat changes and one for a phase change.

The first heat change going from from H2O starting at 88.53 ^oC to 100 ^oC using the formula q=mc delta t - there then is a calculation for a PHASE change at 100 degrees Celsius where liquid water is going through a phase change to steam using delta H vaporization in KJ/mol, then there is the second heat change from 100 deg. Celsius to 124.67 deg. Celsius again using the formula q=mc delta t

step 2)

First heat change calculation:

q=(1.96)(75.3)(373K-361.53K)

= 1692.83J

step 3) calculate the KJ during the phase change from liquid water to steam:

1.96 mole H2O x (heat of vaporization) 40.67KJ/mole = 79.7132 KJ

step 4) calculate the second heat change from steam at 100 degrees Celsius to 124.677 deg. Celsius.

q= (1.96)(35.8)(397.67K - 373K) = 1731.044 J

step 4) add all the heat values with the same units:

1692.83J + 79713.2J + 1731.044J = 83137J or 83.137KJ

Answer 2

to answer the question we need to figure out three things:

1. How much heat is required to raise the temperature of 35.21g water from 88.53C to it's boiling point, 100C
2. How much heat is required to turn 35.21g of water into steam
3. How much heat is required to raise 35.21g of steam from 100C to 125.67C.

We add these three up, we'll get the answer.

So to calculate how much heat is required to change the temperature of something we need the specific heat capacity. Specific heat capacity is a measure of how much heat energy is needed to raise a specific amount of something by a specific temperature. In this instance we are given the specific heat of water in joules (a unit of energy) per mols (a specific number of molecules) per degree kelvin.

Unfortunately, these units don't match the units we were given. The amount of water is in grams, not moles and we're given degrees Celsius and not kelvin. Fortunately 1 degree Celsius = 1 degree Kelvin. But we do need to convert 35.21g water into mols. For this we'll need to molecular mass of water, which is 18.01528g/mol. Now if you're only given a periodic table and can't just google the molecular mass of water, this calculation is a bit harder: Water is 2 hydrogens with an atomic mass of 1.008g/mol each and 1 oxygen has a molecular mass of 16.00g/mol oxygen. 2*1.008+16.00 = 18.015g/mol.

With the molecular mass of water, we can convert grams water into mols water:

35.21g ÷ 18.015g/mol = 1.95 mols

So we have 1.95 mol water that we need to raise from 88.53 C to 100 C. 100° - 88. 53° = 11.47°

So we multiple the amount of water (1.95 mol) times the specific heat of water 75.3J/mol*K times the temperature change (11.47K ) and we get:

1.95 mol * 75.3 J/mol*K * 11.47 K and we get 16841 Joules, which is probably easier to write as 16.841 kJ.

So on to part two, now that the water is at it's boiling point, we can use the heat of vaporization to calculate how much energy is required to break all those hydrogen bonds and turn all that water into steam. Heat of vaporization, by the way is how much energy it takes to turn a specific amount of a substance from a liquid into gas. In this instance it takes 40.67kJ or 4067 Joules to turn 1 mol of water into steam.

Now, we already calculated how many mols of water we're working with - 1.95 mol water, so that saves us a step. We just need to multiple 1.95mol * 40.67kJ/mol = 79.35 kJ.

Part 3: We've converted our 35.21g or 1.95 mol of water into steam, but we need to raise the temperature of that steam from 100C (373.15K) to 125.67C (398.82K). That's an increase of 25.67°C or 25.67 K (1°C = 1K, the scales just have a different zero point).

We set this up like the part 1. We multiply the amount of steam (1.95mol) times the specific heat of steam (35.8 J/mol*K) times the temperature change (25.67K):

1.95mol * 35.8 J/mol*K * 25.67K = 1792.02 joules or 1.792 kJ.

To arrive at the final answer, we just add all the values up:

16.84 kJ to turn bring water to a boil + 79.35 kJ to turn the water into steam + 1.79kJ to turn bring the team to 125.67C

16.84kJ +79.35 kJ +1.79 kJ = 97.98 kJ

And assuming I didn't make any math errors and we aren't being a stickler for significant figures, that's your answer.

A couple things I thought I'd point out:

1. most of the energy goes into turning water into a gas. It takes 100 calories to bring 1g of water from freezing to boiling but 540 calories to turn that water into steam. This make sense when you consider you can get a pot of water to boil in about 10 minutes, but it would take hours before all the water boiled away.
2. The problem is actually phrased in accurately. The specific heat capacity of water is 1 calorie per gram or 4.18 joules per gram as specific heat capacity is defined as the energy to increase the temperature of a substance per unit mass. This problem gave you the molar heat capacity, which is the energy required to increase the temperature of 1 mol of a substance, hence the added step of converting grams of water into mols of water.
3. Always keep track of your units and make sure units cancel. It will prevent errors for example if we multiplied 35.21g water * 18.01g/mol (the molecular mass of water). We'd get 634.1g²/mol which is a meaningless unit. but when we divide 35.21g ÷ 18.01 g/mol we get 1.95 mol -- the units we were expecting.

Hope this helps. Sorry if was too in depth.