Based on the thermodynamic properties provided for water, determine the amount of energy needed for 3.20 kg of water to go from -5.00 °C to 82.0 °C.

Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 3.20 kg of water to go from -5.00 °C to 82.0 °C.

Properties:

Property | Value | Units

Melting Pt. 0.0 C

Boiling Pt. 100.0 C

ΔHfus 6.01 kJ/mol

ΔHvap 40.67 kJ/mol

C_p(s) 37.1 J/mol x C

C_p(l) 75.3 J/mol x C

C_p(g) 33.6 J/mol x C

Answer 1

Nothing wrong with the previous answer, but thought I'd present something a little shorter and perhaps easier to follow.

Since all the thermodynamic constants are given on a per mole basis, we will first convert 3.20 kg of H2O to moles using a molar mass for H2O of 18.0 g / mol (3 sig. figs.)

3.20 kg x 1000 g / kg x 1 mol / 18.0 g = 177.8 moles

Step 1: heat needed to raise 177.8 moles water from -5.00º to 0ºC

q = mC∆T where q = heat = ?; m = mass = 177.8 moles; C = sp. heat of ice = 37.1 J/molº; ∆T = 5º

q = (177.8)(37.1)(5) = 32982 J = 33.0 kJ

Step 2: heat needed to melt 177.8 mols ice @ 0º

q = m∆Hf where q = heat = ?; ∆Hf = heat of fusion = 6.01 kJ/mol

q = (177.8)(6.01) = 1069 kJ

Step 3: heat needed to raise temp of 177.8 mol water from 0º to 82.0º

q = mC∆T where q = heat = ?; m = mass = 177.8 mols; C = sp.heat of liquid = 75.3 J/molº; ∆T = 82.0º

q = (177.8)(75.3)(82.0) = 1097844 J = 1098 kJ

Step 4: Sum up the heats from each step to find total heat needed:

q = 33.0 kJ + 1069 kJ + 1098 kJ

q = 2200.0 kJ

Answer 2

Hey there! Thermo can be tough, but I think it's easier when you look at the units first and use them to make a plan of attack. Additionally, chemistry is a story- it's not just a bunch of numbers, so make sure you don't lose the meaning behind the numbers :)

NOTE: I write HOH instead of H₂O because the subscript is annoying to use.

First thing to ask in these types of problems is, how many mol of compound (water) are we dealing with?

In order to do this, check the molar mass of H, multiply that number by the subscript to the right of the element, and add it to the molar mass of O. If you need more sig-figs than I give you, just follow my procedure with the numbers you need- this sort of thing can vary from instructor to instructor, after all. I will also use approximations. I'm assuming you only need values to one significant figure based on the given values.

H = ~1.0g / mol

H₂=~2.0 g / mol

O = ~15.9 g / mol

H₂O = ~17.9g / mol

We are told there are 3.20kg of HOH for which we must account. Since we have molar mass in g/mol, we have to convert kg to g before converting mass to mol. Remember: since 17.9g HOH = 1 mol HOH you can use the reciprocal of 17.9g / mol and it is still 100% true!

3.2kg HOH * 1000g / kg = 3200g HOH

3200g HOH * 1 mol HOH / 17.9g HOH = 178.77 mol HOH

Rad! Now we have a unit that is in common with a unit in the denominator of many of the given values for Enthalpy (ΔH) and Heat Capacity (Cp(x)), so we can now cancel that unit to simplify these values into JOULES (J)- the unit of energy!

Let's tell the story of this particular quantity of water in small chunks. The water begins at -5ºC in a solid phase and increases heat in that same phase until 0ºC. Okay! So, in Thermo language, we can write this as:

T_i = -5ºC

T_f = 0ºC

ΔT = T_f - T_i (This can be read in English as "the change in temperature is the final temperature minus the initial temperature")

ΔT = 0ºC - (-5ºC)

ΔT = 0ºC + 5ºC

ΔT = 5ºC

And we want to know how much energy that took. What do I know? I know 178.77 mol HOH, and I know ΔT = 5ºC. That means I can use these to units to cancel out similar units! Hmmm...it appears that those Heat Capacity units have mol and ºC in their denominator...AND JOULES IN THE NUMERATOR! Awesome, so, since we were in a solid phase the entire time the temperature was changing, I can do this:

Cp(s) = 37.1 J / (mol*ºC)

37.1 J / (mol*ºC) * 178.77mol = 6632.402235 J / ºC

6632.402235 J / ºC * 5ºC = 33162.01117 J

to represent the energy required to heat solid water by 5ºC

Set this value aside, and let's continue our story!

At 0ºC, we are going to change phase from solid to liquid. This requires us to convert the heat of fusion (ΔHfus) into Joules! Okay, easy!

ΔHfus = 6.01 kJ/mol

6.01 kJ/mol * 178.77mol = 1074.4077 kJ

Since 1 kJ = 1000 J, 1kJ * 1000 J / kJ = 1000 J; we should multiply this number by 1000 to make our energy units agree.

1074.4077 kJ * 1000 J / kJ = 1074407.7J

this represents the amount of energy to convert our quantity of water from solid to liquid.

Let's finish the story! Now the water is a liquid, and it will rise by 82°C in this state to it's final temperature. Let's use the heat capacity of liquid and convert to energy (J) in the same fashion as before. If you get confused here, take some time to look at the given units and the process for our 5 degree temperature rise in the solid state again.

Cp(l) = 75.3 J / (mol * °C)

75.3 J / (mol * °C) * 178.77 mol * 82°C = 1103833.242 J

this number represents the amount of energy to raise the temperature of our quantity of water by 82 degrees in the liquid state.

Our total energy is the sum of all of our "Joules".

ΔE_tot = 33162.01117 J + 1074407.7J + 1103833.242 J

ΔE_{tot =} 2211402.953 J

OR, if your instructor prefers kJ--

2211.40 kJ

Hope this helps! For more comprehensive assistance in chemistry, don't hesitate to contact me through our profile :)