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Question 5, chap 120, sect 6. part 1 of 1 10 points 3 kg of ice at 0Â°C is converted to steam at 100Â°C. The heat of fusion of ice is 79500 cal/kg, the heat of vaporization of water is 540000 cal/kg and its special heat is 1000 cal/kg.Â°C. How much heat is needed? Answer in units of kcal Question 6, chap 120, sect 6. part 1 of 1 10 points You have 56 g of steam at 100Â°C. The specific heat capacity of water is 4180 J/kg. C and latent heat of vaporiza- tion of water is 2.26 x 106 J/kg How much heat must be removed to change it to 56 g of water at 25Â°C? Answer in units of J Question 7, chap 120, sect 9. part 1 of 1 10 points Atmospheric pressure is 101000 Pa A toy balloon is inflated with helium at a constant pressure that is 790000 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 0.0003 m3 to 0.00184 m3, how much work is done by the balloon on the surrounding air? Answer in units of J Solution 5) Q = Q (to melt ice) + Q (to take water from 0 to 100 oC) + Q (to vapourize water) = m*Hfus + m*s*(Tf-Ti) + m*Hvap = 3*79500 + 3*1000*(100-0) + 3*540000 = 3.2385*10^6 cal = 32385 Kcal Answer: 32385 Kcal 7) As the volume of the balloon expands, it must do work against the atmosphere, this is the work which must be calculated. Remember work done = Force x distance and that Pressure = Force/Area rearranging the second equation will give Force = P x A, plug this into the first equation Work Done = P x A x s = P (A x s) = P x V = pressure x change in volume so for our question, work done = (0.79 Ã— 10^6) x (0.00184-0.0003) = 1216 .6 Joules .

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Question 5, chap 120, sect 6. part 1 of 1 10 points 3 kg of ice at 0Â°C is converted to steam at 100Â°C. The heat of fusion of ice is 79500 cal/kg, the heat of vaporization of water is 540000 cal/kg and its special heat is 1000 cal/kg.Â°C. How much heat is needed? Answer in units of kcal Question 6, chap 120, sect 6. part 1 of 1 10 points You have 56 g of steam at 100Â°C. The specific heat capacity of water is 4180 J/kg. C and latent heat of vaporiza- tion of water is 2.26 x 106 J/kg How much heat must be removed to change it to 56 g of water at 25Â°C? Answer in units of J Question 7, chap 120, sect 9. part 1 of 1 10 points Atmospheric pressure is 101000 Pa A toy balloon is inflated with helium at a constant pressure that is 790000 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 0.0003 m3 to 0.00184 m3, how much work is done by the balloon on the surrounding air? Answer in units of J Solution 5) Q = Q (to melt ice) + Q (to take water from 0 to 100 oC) + Q (to vapourize water) = m*Hfus + m*s*(Tf-Ti) + m*Hvap = 3*79500 + 3*1000*(100-0) + 3*540000 = 3.2385*10^6 cal = 32385 Kcal Answer: 32385 Kcal 7) As the volume of the balloon expands, it must do work against the atmosphere, this is the work which must be calculated. Remember work done = Force x distance and that Pressure = Force/Area rearranging the second equation will give Force = P x A, plug this into the first equation Work Done = P x A x s = P (A x s) = P x V = pressure x change in volume so for our question, work done = (0.79 Ã— 10^6) x (0.00184-0.0003) = 1216 .6 Joules .

- 1. Question 5, chap 120, sect 6. part 1 of 1 10 points 3 kg of ice at 0Â°C is converted to steam at 100Â°C. The heat of fusion of ice is 79500 cal/kg, the heat of vaporization of water is 540000 cal/kg and its special heat is 1000 cal/kg.Â°C. How much heat is needed? Answer in units of kcal Question 6, chap 120, sect 6. part 1 of 1 10 points You have 56 g of steam at 100Â°C. The specific heat capacity of water is 4180 J/kg. C and latent heat of vaporiza- tion of water is 2.26 x 106 J/kg How much heat must be removed to change it to 56 g of water at 25Â°C? Answer in units of J Question 7, chap 120, sect 9. part 1 of 1 10 points Atmospheric pressure is 101000 Pa A toy balloon is inflated with helium at a constant pressure that is 790000 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 0.0003 m3 to 0.00184 m3, how much work is done by the balloon on the surrounding air? Answer in units of J Solution 5) Q = Q (to melt ice) + Q (to take water from 0 to 100 oC) + Q (to vapourize water) = m*Hfus + m*s*(Tf-Ti) + m*Hvap = 3*79500 + 3*1000*(100-0) + 3*540000 = 3.2385*10^6 cal = 32385 Kcal Answer: 32385 Kcal 7) As the volume of the balloon expands, it must do work against the atmosphere, this is the work which must be calculated. Remember work done = Force x distance and that Pressure = Force/Area rearranging the second equation will give Force = P x A, plug this into the first equation Work Done = P x A x s = P (A x s) = P x V = pressure x change in volume so for our question, work done = (0.79 Ã— 10^6) x (0.00184-0.0003) = 1216 .6 Joules