• 100% Satisfaction Guarantee

mba-sci, instructor
Category: Homework
Satisfied Customers: 205
Experience:  mba degree, m.sc. degree in chem. eng., biz consultant
4876183
mba-sci is online now

# 1. How many calories of heat are required to change 1

### Resolved Question:

1.      How many calories of heat are required to change 1 gram of ice at 0ºC to liquid water at 0ºC?

2.      How many calories of heat are required to change 1 gram of water at 100ºC to 1 gram of steam at 100ºC?

3.      How many calories are needed to melt 50 grams of ice at 0ºC and convert it to liquid water at 35ºC?

4.      For the irreversible transition of: 100 g H2O (s, -25ºC) ? 100 g H2O (v, +120ºC) Calculate the total amount of heat necessary to effect a set of reversible reactions for theses conditions.

5.      In a steam radiator, steam at 110ºC condenses and the liquid water formed is cooled to 80ºC. How much heat per gram of liquid formed, is released in this process?

6.      How much heat is absorbed by a person's hand if 100 grams of liquid water at 100ºC is poured on his hand? (Assume the final temperature of the water will be normal body temperature of 37ºC)

7.      How much heat is absorbed by a person's hand if 100 grams of steam at 100ºC is first converted to liquid water at 100ºC and then cooled to normal body temperature?

8.      Which causes the more severe burn hot water or steam? Why?
Submitted: 8 years ago.
Category: Homework
Expert:  mba-sci replied 8 years ago.

(More answeres to come up ASAP.)

4. For the irreversible transition of: 100 g H2O (s, -25ºC) ? 100 g H2O (v, +120ºC) Calculate the total amount of heat necessary to effect a set of reversible reactions for theses conditions.

The specific heat of water (ICE) is 0.504 calorie/gram °C. So for 100 grams of solid water (ice) at -25ºC) to change to 0 degrees, the heat needed is 0.504x100x(0-(-25))=1260 (cal).

For ice, the latent heat (from solid water to liquid water) is 80 calories per gram. So the heat needed when 100 grams of ice at 0 degrees celsius are converted to liquid water at 0 degrees is 80x100=8000 (cal).

The specific heat of water is 1 calorie/gram °C. So for 100 grams of liquid water at 0 degrees celsius to be heated to 100ºC (liquid water), the heat needed is 1x100x(100-0)=10000 (cal).

For water, the heat of vaporization is 540 calories per gram. So the heat needed when 100 grams of liquid water at 100 degrees celsius are heated to steam water vapor at 100 degrees is 540x100=54000 (cal).

The specific heat of water VAPOR is 0.477 calorie/gram °C. So for 100 grams of vapor water at 100 degrees celsius to be heated to 120 degrees, the heat needed is 0.477x100x(120-100)=954 (cal).

Hence the heat needed is 1260+8000+10000+54000+954 =74214(cal).

5. In a steam radiator, steam at 110ºC condenses and the liquid water formed is cooled to 80ºC. How much heat per gram of liquid formed, is released in this process?

The specific heat of water VAPOR is 0.477 calorie/gram °C. So for 1 gram of liquid water at 110 degrees celsius to be cooled to 100 degrees, the heat released should be 0.477x1x(110-100)=4.77 (cal).

For water, the heat of vaporization is 540 calories per gram. So the heat released when 1 gram of steam at 100 degrees celsius are converted to liquid water at 100 degrees should be 540x1=5400 (cal).

The specific heat of water is 1 calorie/gram °C. So for 1 gram of liquid water at 100 degrees celsius to be cooled to 80ºC the heat released should be 1x1x(100-80)=20 (cal).

Hence the heat per gram released in this process is 4.77+5400+20=5424.77(cal).

7. How much heat is absorbed by a person's hand if 100 grams of steam at 100ºC is first converted to liquid water at 100ºC and then cooled to normal body temperature?

For water, the heat of vaporization is 540 calories per gram. This means that the heat released when 100 grams's of steam at 100 degrees celsius are converted to liquid water at 100 degrees should be 540x100=540000 (cal).

We know that the specific heat of water is 1 calorie/gram °C. So for 100 grams of liquid water at 100 degrees celsius to be cooled to normal body temperature (37oC), the heat released should be 1x100x(100-37)=6300 (cal).

Hence the heat absorbed by a person's hand is 540000+6300=546300(cal).

Customer: replied 8 years ago.
Reply to mbasci's Post: I will wait until you have finished and yes I will gladly give a bonus.Customer
Expert:  mba-sci replied 8 years ago.
You can view this answer by clicking here to Register or Login and paying \$3.
mba-sci, instructor
Category: Homework
Satisfied Customers: 205
Experience: mba degree, m.sc. degree in chem. eng., biz consultant

Ask-a-doc Web sites: If you've got a quick question, you can try to get an answer from sites that say they have various specialists on hand to give quick answers... Justanswer.com.
...leave nothing to chance.
Traffic on JustAnswer rose 14 percent...and had nearly 400,000 page views in 30 days...inquiries related to stress, high blood pressure, drinking and heart pain jumped 33 percent.
Tory Johnson, GMA Workplace Contributor, discusses work-from-home jobs, such as JustAnswer in which verified Experts answer people’s questions.
I will tell you that...the things you have to go through to be an Expert are quite rigorous.

### What Customers are Saying:

• Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C. Freshfield, Liverpool, UK
< Previous | Next >
• Wonderful service, prompt, efficient, and accurate. Couldn't have asked for more. I cannot thank you enough for your help. Mary C. Freshfield, Liverpool, UK
• This expert is wonderful. They truly know what they are talking about, and they actually care about you. They really helped put my nerves at ease. Thank you so much!!!! Alex Los Angeles, CA
• Thank you for all your help. It is nice to know that this service is here for people like myself, who need answers fast and are not sure who to consult. GP Hesperia, CA
• I couldn't be more satisfied! This is the site I will always come to when I need a second opinion. Justin Kernersville, NC
• Just let me say that this encounter has been entirely professional and most helpful. I liked that I could ask additional questions and get answered in a very short turn around. Esther Woodstock, NY
• Thank you so much for taking your time and knowledge to support my concerns. Not only did you answer my questions, you even took it a step further with replying with more pertinent information I needed to know. Robin Elkton, Maryland
• He answered my question promptly and gave me accurate, detailed information. If all of your experts are half as good, you have a great thing going here. Diane Dallas, TX

• ### Manal Elkhoshkhany

#### Satisfied Customers:

4530
More than 5000 online tutoring sessions.
< Last | Next >

### Manal Elkhoshkhany

#### Satisfied Customers:

4530
More than 5000 online tutoring sessions.

### LogicPro

#### Satisfied Customers:

4395
Expert in Java C++ C C# VB Javascript Design SQL HTML

### Linda_us

#### Satisfied Customers:

3132
Post Graduate Diploma in Management (MBA)

### Chris M.

#### Satisfied Customers:

2486
Master's Degree, strong math and writing skills, experience in one-on-one tutoring (college English)

### F. Naz

#### Satisfied Customers:

2055
Experience with chartered accountancy

### Bizhelp

#### Satisfied Customers:

1884
Bachelors Degree and CPA with Accounting work experience