The next concept to be considered when dealing with heat, is “Latent Heat”. The

Question

The next concept to be considered when dealing with heat, is “Latent Heat”. The word latent refers to the temperature of the system remaining the same even as heat is being added or subtracted from it. This will occur when the system is changing states. Consider a piece of ice.

9. The latent heat of fusion (or the amount of heat required to turn the ice into water) is 80 Calories/gram. How much heat must be added to melt 2 grams of ice at 0°C and what is the final temperature after this heat is added? _____________   _____________ (3 pts)

10. The latent heat of vaporization (or the amount of heat required to turn the water into a vapor) is 540 Calories/gram. How much heat must be added to turn 3 grams of water at 100°C into a vapor and what will be the final temperature after this much heat is added? _________   _________ (3 pts)

Putting sensible and latent heat together, it is now possible to figure out the temperature of water going through various stages of heat input/withdrawal. (Heat capacity of water is 1 cal / g °C)

11. 60 calories is added to 1 gram of water at an initial temperature of 80°C. What is the water’s final temperature and state (solid, liquid or vapor)?

Temperature = ______________ (2 pts)

State = ____________________ (1 pt)

12. 120 calories is removed from 1 gram of water at an initial temperature of 40°C. What is the water’s final temperature and state (solid, liquid or vapor)?

Temperature = _______________ (2 pts)

State = ____________________ (1 pt)

140 120 100 Latent Latent heat of vaporization (540 cal/gram) 0 Water vapor heat of 80 fusion 60 (80 cal/ gram) Liquid water 2010 -20lce 20 100 200 400 600 780 800 Heat (calories/gram)

Explanation / Answer

Question 9 heat required to melt 2 g of water is 80x2= 160 calories.The final temeprature at the point when all ice has turned into water will be 0 degrees centigrade.

Question 10 The heat required will be 540 x 3 = 1620 calories The final temeprature at the point when all water has turned into water pour will be 100 degrees centigrade.

Question 11 Let the final temp be 100 ( it cannot go beyond 100 ) to heat gained in cals we can solve as follows

(mass of waterx heat capacity x diff in temp) + ( calories reqd for change of state) = amount of calories

amount of cals reqd to heat to 100 degrees =1x1 x 20 = 20 calories Now remaining calories will be 60 - 20 which is 40 calories. Now 1 g of water requires 540 cal for change of state.Hence 500 cals more are required Hence most of the water will remain in liquid state and the temperature of water will remain at 100 degree centigrade.

So temperature of water will be 100 degree centigrade

Stare will be liquid

Question12 Now 120 calories are removed temp of water is 40 degrees

Now 1 g of water will need 40 calries to coll from 40 gegrees to 0 degrees Hence 120 - 40 = 80 calories will be utilised to freeze

aLL 80 WILL BE USED because latent heat is 80 so all water will feeze

Temperature will be 0 degree centigrade

state ill be solid

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