# Ch 13.2 help

## Ch 13.2 help

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.

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.

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.

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.

How to solve:

periwinkle

Posts : 22
Join date : 2008-09-17

## Re: Ch 13.2 help

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

Guest01

Posts : 133
Join date : 2008-09-19

## Re: Ch 13.2 help

Nevermind, I got Question 4

Guest01

Posts : 133
Join date : 2008-09-19

## number 4

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

student
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hwilson
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## Re: Ch 13.2 help

What have you all gotten to so far?

Guest01

Posts : 133
Join date : 2008-09-19

## Hints

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.

Physics1
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## quiz

Thank you thank you thank you!

student
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## Re: Ch 13.2 help

i got number 4 right but how you got that solution

dec1 010
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## #4

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

super mo
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## Questions with numbers

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.

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.

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

Kathleen
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## Re: Ch 13.2 help

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.

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

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.

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

Kathleen
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## Re: Ch 13.2 help

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

asfgbdfb
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## Re: Ch 13.2 help

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

Guest01

Posts : 133
Join date : 2008-09-19