Ch 13.2 help
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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 g1 oC1 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 g1 oC1 for the specific heat of water vapor and 540 cal g1 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:
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 g1 oC1 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 g1 oC1 for the specific heat of water vapor and 540 cal g1 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 : 20080917
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
I have all the others which I will post up How To... shortly
Guest01 Posts : 133
Join date : 20080919
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 g1 oC1 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 97.3244.671568 = 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 g1 oC1 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 97.3244.671568 = 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 g1 oC1 for the specific heat of water vapor and 540 cal g1 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)(269.5100) / (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 g1 oC1 for the specific heat of water vapor and 540 cal g1 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)(269.5100) / (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 : 20080919
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