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

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?

(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 9 years ago.
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