Problems 4 - 6 refer to the following: an ideal gas initially at temperature Ta
ID: 1397955 • Letter: P
Question
Problems 4 - 6 refer to the following: an ideal gas initially at temperature Ta = 280 K, with CV=3/2R and CP=5/2R undergoes a transformation as shown in the figure.
4. The temperature in state b is Tb:
A) 70 K
B) 93 K
C) 280 K
D) 840 K.
E) 1120 K
5. The heat absorbed in the process (in MJ=106J) is
A) 22.5 MJ
B) 13.5 MJ
C) 9.0 MJ
D) 2.25 MJ
E) 0.9 MJ
6. The percentage from heat that is transformed into work (W/Q x100) is:
A) 1%
B) 4 %
C) 40 %
D) 96 %
E) 100%
7. If the heat required to raise the temperature of an object from 5 degrees C to 10 degrees C is Q, the heat required to increase the temperature from 10 degrees C to 15 degrees C is:
A) Q/3
B) Q/2
C) Q
D) 3Q/2
E) 2Q
8. How much water at 10 degrees C (cW=4.2 kJ/kgoC ) is needed to melt 100g ice at 0 degrees C. (L = 334 kJ/kg)
A) 0.79 Kg
B) 3.34 Kg
C) 2.8 Kg
D) 7.95 Kg.
E) 1.12 Kg
9. When the temperature of an ideal gas changes from 200K to 300K while the volume remain constant, its pressure changes from 0.6 atm to
A) 0.2 atm.
B) 0.3 atm.
C) 0.4 atm
D) 0.8 atm
E) 0.9 atm
F) 1.2 atm
Explanation / Answer
here,
mass of iron , mi = 50 g
mi = 0.05 kg
specific heat of iron , Cfe = 480 J/kg degree C
temprature of iron , Tfe = 200 degree C
mass of water , mw = 200 g
mw = 0.2 kg
specific heat of water , Cw = 4200 J/kg degree C
temprature of water , Tw = 200 degree C
let the final temprature be Tf
heat lost by iron = heat gain by water
mi * Cfe * (Tfe - Tf) = mw * Cw * (Tf - Tw)
0.05 * 480 * (200 - Tf) = 0.2 * 4200 * (Tf - 20)
Tf = 25 degree C
the final temprature of water is (C) 25 degree C
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