Question

Heat absorption problems

ONE:
"Calculate the total amount of heat absorbed (in KJ) when 2.00 mol of ice at -30.0 °C is converted to steam at 140.0 °C"

• 2.00 mol ice = 36.04 g ice
• Cp (ice) = 2.06 J/g °C
• Cp (water) = 4.18 J/g °C
• Cp (steam) = 1.87 J/g °C
• ∆Hfus = 6.01 kJ/mol
• ∆Hvap = 40.7 kJ/mol
• ∆Htotal = ? kJ

TWO:
"What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?"

Answer 1

the problem ask how much heat you have to suministrarte to a piece of ice at -30C to became vapor at 140C


1) raise 2.00 mol (36.04g) of ice from -30.0 to zero Celsius:

(36.04 g) (30.0 C) (2.06 J g¯1 K¯1) = ∆H1 J

2) melt 36.04 g of ice:

(36.04g) (6.01 kJ/mol) = ∆H2 J

3) raise 36.04 g of liquid water from zero to 100.0 Celsius:

(36.04g) (100.0 K) (4.184 J g¯1 K¯1) = ∆H3 J

4) evaporate 36.04 g of liquid:

(36.04 g) (∆Hvap) = ∆H4 J

5) raise 36.04 g of steam from 100.0 to 140.0 Celsius:

(36.04g) (40.0 C) (Cp (steam) ) = ∆H5 J
6) add the results:

∆H1 + ∆H2 + ∆H3 + ∆H4+ ∆H5= ∆Htotal kJ

Answer 2

do you know what is heat?

Answer 3

I don't know what "suministarte" is ;-;

Answer 4

supply

Answer 5

What ∆ value do I use in the first part, to get ice to water (0 °C?)

... if I'm going with the right idea here idk

Answer 6

when you put a kettle on the stove to heat up water to make a tea you have to supply a certain amount of heat to the water to boil it. If you have to make two cup of tea you need more water and you need more heat to heat up the water for two cups of tea

Answer 7

Okay.
What ∆ value is used for the ice → water conversion?

Answer 8

first you have to go from ice at -30C to ice at 0C

Answer 9

then from ice at 0C to water at 0C

Answer 10

∆Hfus = 6.01 kJ/mol ice to water conversion

Answer 11

Okay
So then
ice -30 to ice 0
ice 0 to water 0
water 0 to water 100
water 100 to steam 100?

Answer 12

and steam 100 to steam 140 C

Answer 13

Okay
so the ∆ for ice -30 to ice 0 is ... ?
(Sorry, I got a little confused)

Answer 14

Cp (ice) = 2.06 J/g °C and you have to multiply by the mass of ice that you have and the ∆ temperature. This is the heat for 1 gram of ice to change 1 C

Answer 15

Alright thanks

So any ice temp changes uses the Cp for ice... and if it changes states then which Cp (or whatever ∆ value) would I use?

Answer 16

What I understand so far regarding temperature changes:
ice → ice uses Cp for ice
water → water uses Cp for water
steam → steam uses Cp for steam

Would ∆H-fusion and ∆H-vap count for state changes?

Answer 17

Would ∆H-fusion and ∆H-vap count for state changes?
yes use them for state change a constant temperature. you multiply only by the mass

Answer 18

Ah, yeah, sorry! I realized you'd already expounded on that in your first comment. Apologies for making you repeat things ;-;

Answer 19

Okay, then I think I've got a basic grasp on everything.

er... may I ask a question about molarity?

Answer 20

Actually, it's a solution problem oops

"A solution is prepared with 0.30 g of salt and 30.0 mL of water. What is the solute, and what is the solvent?"

Answer 21

the salt is the solute and the water is the solvent

Answer 22

Oh.
May I know why? (Just for future reference) o: