Please help !!! 4. Five kilograms of water at 100°C is converted into steam. The
ID: 1574610 • Letter: P
Question
Please help !!! 4. Five kilograms of water at 100°C is converted into steam. The heat added to the water is (Lt 333 kJ/kg, Lv- 2256 kJ/kg) a. 1,665 kJ b. 11.3 X 103 kJ c. 33.3 kJ d. 1.13 X 106 kJ 5. How much heat must be removed from 2 kg of water at 20°C to convert it into ice at 0°C? (cw 4190 J/(kg-K, Lf (water) 333,000 J/kg) a. 167.6 k b. 666 kJ c. 834 kJ d. 2000 kJ 6. When heat is added to boiling water, its temperature a. increases b. decreases c. stays the same ·The specific heat of lead is 128 J/(kg-K). If 0.3 kg of lead at 100°C is mixed with 0.I kg of water (Qw 4190 J/kg-K) at 70"C in an insulated container, the final temperature of the mixture is a. 100°C b. 85.5 c. 79.5°C d. 72.5 . An ideal gas is maintained at a constant pressure of 70,000 Pa during an isobaric process and its volume increases by 0.2 m. How much work is done by the gas? a. -14,0000 J b. 14,000 J c. 35,000 J d. -35,000 Five moles of an ideal gas expands isothermally at 97°C to three times its initial volume. The work done by the gas is (R 8.31 J/mole-K) a. 1.7 X 104J b. 4.4 X 103 J c. 1.9 X 10. d. 1.0 X 104JExplanation / Answer
4) Heat required is the latent heat of vapourization = mLv = 5*2256 = 11.3*103 kJ: option b is correct
5) heat removed, Q= mcw(T2-T1) + mLf = 5*(4.19*(293-273) + 333)
Q= 2084 kJ option d is correct
6) option c stays the same. Because temperature remains constant during phase constant.
7) heat lost by lead = heat gained by water
let the final temperature be T
0.3*128*(373-T)= 0.1*4190*(T-70)
T=345.51 K or T=72.51 option d is correct.
8) W= P*dV =70000*(0.2) = 14000 J
option b is correct
9) During isothermal expansion internal energy is zero, hence
Q= W = p dV
p=nRT/V
W= nRT dV/V
W=nRT*ln(V2/V1)
W= 5*8.31*(273+97)*ln(5)
W=1.9*104 J approx option c is correct