mloves_football
New member
1.) Consider the problem of melting ice in a liquid (water), . of the amount of ice is 50 g at 0 C , and the amount of water is 50g at 25 C, what is the final temperature in degrees Centigrade?
I know that to fin the answer it is set up like:
(Heat to change .05 kg of ice to water)+ (heat to raise .05 kg of water from 0 C to T) = (heat lost by .05 kg of water cooling from 25 C to T)
(.05kg)(80 kcal/kg) + (.05 kg) (1 kcal/kg C)(T-0C)=(0.05 kg)1 kcal/kgC)(25 C- T)
2.) A colorimeter (mc= 0.03 Kcal/ c) contains 270 g of water at 10 C. onto this calorimeter 200g of ice at -50 C is dropped; then 10 g of steam at 120 C is introduced. What is the final temperature of the system in C?
C steam= C ice= 0.5 kcal/kg C
the heat of vaporation is for steam is 540 kcal/kg
I know that to fin the answer it is set up like:
(Heat to change .05 kg of ice to water)+ (heat to raise .05 kg of water from 0 C to T) = (heat lost by .05 kg of water cooling from 25 C to T)
(.05kg)(80 kcal/kg) + (.05 kg) (1 kcal/kg C)(T-0C)=(0.05 kg)1 kcal/kgC)(25 C- T)
2.) A colorimeter (mc= 0.03 Kcal/ c) contains 270 g of water at 10 C. onto this calorimeter 200g of ice at -50 C is dropped; then 10 g of steam at 120 C is introduced. What is the final temperature of the system in C?
C steam= C ice= 0.5 kcal/kg C
the heat of vaporation is for steam is 540 kcal/kg