# 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.

__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:__

**periwinkle**- Posts : 22

Join date : 2008-09-17

## Re: Ch 13.2 help

Has anyone gotten

I have all the others which I will post up

*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

## 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.

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**- Guest

## 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.

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

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

**Kathleen**- Guest

## 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.

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

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

**Kathleen**- Guest

## Re: Ch 13.2 help

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

Your help is greatly appreciated

Your help is greatly appreciated

**asfgbdfb**- Guest

## 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

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