Ch 13.2 help

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Ch 13.2 help

Post  periwinkle on Tue Nov 18, 2008 12:11 am

Question 1
53.39 grams of steel at 106.2 oC is put into 0.7382 liter of water at 20 oC. What is the final temperature at equilibrium assuming no heat loss to the environment (see sheet 12; use 0.107 cal g-1 oC-1 for the specific heat of steel) ? Indicate with a positive (negative) sign whether the steel looses less (the same amount of) heat than the water gains.

Answer: -20.66195577268855
How to solve:

Question 2
97.32 grams of ice are added to 148.1 grams of water at a temperature of 24.04 oC. How much ice is left over at equilibrium. (see sheet 14,17,18) (Use 79.7 cal/g for the latent heat of fusion of water) ? Indicate with a positive (negative) sign whether the final temperature is greater than zero (is zero) oC.

Answer: -52.64843161856964
How to solve:

Question 3
31.51 grams of ice are added to 212.4 grams of water at a temperature of 34.53 oC. What is the final temperature at equilibrium (see sheet 18') (Use 79.7 cal/g for the latent heat of fusion of water) ? Indicate with a positive (negative) sign whether all (not all) ice melts.

Answer: 19.772969537944327
How to solve:

Question 4
312.5 grams of steam at 269.5 oC are added to 85.58 grams of water at 100 oC. How much water is left over at equilibrium (see sheet 19'; hint: not evaporating all the water with the hot steam is analogous to not melting all the ice with the hot water in Ch13_1 #6; use 0.5 cal g-1 oC-1 for the specific heat of water vapor and 540 cal g-1 for the latent heat of vaporization ) ? Indicate with a positive (negative) sign whether the final temperature is greater than 100 (is 100) oC.

Answer: -36.53486111111111
How to solve:

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Re: Ch 13.2 help

Post  Guest01 on Sun Nov 23, 2008 7:58 pm

Has anyone gotten Question 4? I know it's nearly the same procedure as Question 2.

I have all the others which I will post up How To... shortly

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Re: Ch 13.2 help

Post  Guest01 on Wed Nov 26, 2008 7:20 pm

Nevermind, I got Question 4

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number 4

Post  student on Fri Nov 28, 2008 12:22 am

Can someone please post some help on number 4? I'm lost

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Question 2 and 3

Post  hwilson on Fri Nov 28, 2008 3:45 pm

Can someone please help with question 2 and 3?

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Re: Ch 13.2 help

Post  Guest01 on Sat Nov 29, 2008 3:34 am

What have you all gotten to so far?

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Hints

Post  Physics1 on Sat Nov 29, 2008 3:26 pm

1 . Tf = ((Ms*c*Ts) + (Liters*1000*Tw)) / ((Ms*c) + (Liters*1000))

make sure it's all in grams.
Negate.

2. Mmelt = (MwCwTw) / L

Then Mice - Mmelt = Magic

Negate.

3. Tf = (MwCwTw - L*Mice) / (MiceCw + MwCw)

Positive.

4. Mvap = ((Msteam*c)(Tsteam - 100) / L -----c is the smaller number given

Then Mwater - Mvap = 4/4 on the Hw

Negate.

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quiz

Post  student on Sun Nov 30, 2008 2:44 pm

Thank you thank you thank you! Very Happy

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Re: Ch 13.2 help

Post  dec1 010 on Mon Dec 01, 2008 3:09 am

i got number 4 right but how you got that solution

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#4

Post  super mo on Mon Dec 01, 2008 12:56 pm

I did not come up with the solution I only went by the post, sorry.

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Questions with numbers

Post  Kathleen on Mon Dec 01, 2008 8:25 pm

Question 1
53.39 grams of steel at 106.2 oC is put into 0.7382 liter of water at 20 oC. What is the final temperature at equilibrium assuming no heat loss to the environment (see sheet 12; use 0.107 cal g-1 oC-1 for the specific heat of steel) ? Indicate with a positive (negative) sign whether the steel looses less (the same amount of) heat than the water gains.

Answer: -20.66195577268855

Using formula given in one of the above posts,
Tf = (53.39*.107*106.2)+(.7382*1000*20)/((53.39*.107)+(.7382*1000))= 15370.69/743.9 = 20.6619 and it is negative

Question 2
97.32 grams of ice are added to 148.1 grams of water at a temperature of 24.04 oC. How much ice is left over at equilibrium. (see sheet 14,17,1 (Use 79.7 cal/g for the latent heat of fusion of water) ? Indicate with a positive (negative) sign whether the final temperature is greater than zero (is zero) oC.

Answer: -52.64843161856964

Mmelt = (148.1 * 1 * 24.04) / (79.7) = 44.671568
Then do mass of ice 97.32-44.671568 = 52.64843
Answer is negative

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Re: Ch 13.2 help

Post  Kathleen on Mon Dec 01, 2008 8:34 pm

Question 3
31.51 grams of ice are added to 212.4 grams of water at a temperature of 34.53 oC. What is the final temperature at equilibrium (see sheet 18') (Use 79.7 cal/g for the latent heat of fusion of water) ? Indicate with a positive (negative) sign whether all (not all) ice melts.

Answer: 19.772969537944327


Tf = (212.4*1*34.53)-(79.7*31.51) / (31.51*1+212.4*1) = (4822.825)/(243.91) = 19.77
Answer is positive


Question 4
312.5 grams of steam at 269.5 oC are added to 85.58 grams of water at 100 oC. How much water is left over at equilibrium (see sheet 19'; hint: not evaporating all the water with the hot steam is analogous to not melting all the ice with the hot water in Ch13_1 #6; use 0.5 cal g-1 oC-1 for the specific heat of water vapor and 540 cal g-1 for the latent heat of vaporization ) ? Indicate with a positive (negative) sign whether the final temperature is greater than 100 (is 100) oC.

Answer: -36.53486111111111

Mvap= (312.5*.5)(269.5-100) / (540) = (26484.375)/540 = 49.045
Mwater - Mvap = 85.58 - 49.045 = 36.53486
Answer is negative

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Re: Ch 13.2 help

Post  asfgbdfb on Wed Dec 03, 2008 2:21 pm

Thank you all for posting the procedure, but could you also post the formulas you used?

Your help is greatly appreciated

asfgbdfb
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Re: Ch 13.2 help

Post  Guest01 on Thu Dec 04, 2008 2:06 am

I'm pretty sure the formulas (for the most part) are posted. They maybe slightly different then the page the question referenced though.

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