1. Question 5, chap 120, sect 6. part 1 of 1 10 points 3 kg of ice at 0°C is converted to steam at
100°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.°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 100°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 25°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 × 10^6) x (0.00184-0.0003) = 1216 .6 Joules