DAT Gen. Chem. PROBLEM????? FREE BEER

[testnein]

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No free beer but I have a question im confused on. Im taking a Kaplan course and as some may know, they just do example problems and give the answers instead of explaing how they got those answers. So here's the problem.
HOW MUCH HEAT MUST BE ADDED TO BRING 20g OF ICE AT 0 DEGREES C TO WATER VAPOR AT 100 DEGREES C? (THE HEAT OF FUSION OF ICE IS 80 cal/g. THE HEAT OF VAPORIZATION IS 540 cal/g.)
Kaplan says the answer is 14400cal, I keep getting 18960 cal
What the **** is going on?

 

[fsddsms]

No free beer but I have a question im confused on. Im taking a Kaplan course and as some may know, they just do example problems and give the answers instead of explaing how they got those answers. So here's the problem.
HOW MUCH HEAT MUST BE ADDED TO BRING 20g OF ICE AT 0 DEGREES C TO WATER VAPOR AT 100 DEGREES C? (THE HEAT OF FUSION OF ICE IS 80 cal/g. THE HEAT OF VAPORIZATION IS 540 cal/g.)
Kaplan says the answer is 14400cal, I keep getting 18960 cal
What the **** is going on?

Q = mc(delta T)
- Take Q for each step from melting to boiling

(20g)(80cal/g) + (20g)(1C/g)(100C) + 20g(540cal/g)
1600 + 2000 + 10,800 cal
= 14,400 cal

Edit: gimme beer!

 

[Dentite99]

No free beer but I have a question im confused on. Im taking a Kaplan course and as some may know, they just do example problems and give the answers instead of explaing how they got those answers. So here's the problem.
HOW MUCH HEAT MUST BE ADDED TO BRING 20g OF ICE AT 0 DEGREES C TO WATER VAPOR AT 100 DEGREES C? (THE HEAT OF FUSION OF ICE IS 80 cal/g. THE HEAT OF VAPORIZATION IS 540 cal/g.)
Kaplan says the answer is 14400cal, I keep getting 18960 cal
What the **** is going on?

For These types of problems I try to draw that temperature/phase change graph.

q1 is the first plateau, which is phase change= m(H)
q2 is the slope, which is temp change= mCdeltaT
q3 is the 2nd plateau, which is phase change=mCdeltaT

q1= (20)(80)=1600
q2=(20)(1)(100)=2000
q3=(20)(540)=10800

1600+2000+10800=14400cal

I can never do these in my head, that's why the graph helps.

 

[fsddsms]

For These types of problems I try to draw that temperature/phase change graph.

q1 is the first plateau, which is phase change= m(H)
q2 is the slope, which is temp change= mCdeltaT
q3 is the 2nd plateau, which is phase change=mCdeltaT

q1= (20)(80)=1600
q2=(20)(1)(100)=2000
q3=(20)(540)=10800

1600+2000+10800=14400J

I can never do these in my head, that's why the graph helps.

^^ yup! make sure to note that it's not Joules though. Question is in calories.

 

[testnein]

Sweet that’s how I approached the problem, but I was using 4.18 as the S.H. for water not 1. Where the hell did I get 4.18 for the S.H of water? O wait thats in J, yep I feel stupid. Thanks guys.

 

[fsddsms]

Sweet that’s how I approached the problem, but I was using 4.18 as the S.H. for water not 1. Where the hell did I get 4.18 for the S.H of water? O wait thats in J, yep I feel stupid. Thanks guys.

U did it correctly. 4.18 JOULES / degree celcius vs 1 Cal / gram x degree C just mixed which one to use to match with the units the question is asking. Keep both numbers to memory though. 👍