A cylinder contains 0.5 kg of steam at the temperature of 100 degree C trapped b
ID: 1024019 • Letter: A
Question
A cylinder contains 0.5 kg of steam at the temperature of 100 degree C trapped below a piston. The piston is 1.3 kg and its area is 10 cm^2. The pressure above the piston is 100 kPa. The steam is heated to 102 degree C. Assuming that the value of C_p for steam remains 1.89 kJ kg^-1 K^-1 during the temperature range, discuss the total internal energy change of steam in each case. (g=9.8 m s^-2, 0 degree C=273.15 K) The piston is slowly elevated at the height of 20 cm (constant pressure inside the piston). The piston is held in place by latches.Explanation / Answer
Closed system so:
Q - W = dU + dEk + dEp
ignore Ek and Ep since there is no change or too small to consider
Q - W = dU
Q at constant pressure:
Q = m*C*dT
Q = 0.5 kg * 1.89 kJ/kgK * (102-100 C)
Q = 1.89 kJ
For work:
W = P*dV = P*A*dh
P = Ppiston + Padded
A = 10 cm2 = 10 cm2 /(100^2 cm2 /m2) = 0.001 m2
Ppiston = Wpiston/A = Mpiston*g/A = 1.3 kg*9.8 m/s2 /(0.001 m2 ) = 12740 Pa
Padded = 100 kPa = 100000 Pa
P = Ppiston + Padded = 12740 +100000 = 112740 Pa = 112.740 kPa
W = P*dV = P*A*dh
dh = 20 cm = 0.2 m
W = P*dV = 112.740 kPa * 0.001 m2* (0.2 m) = 0.022548 kJ
W is being lost by system so, it must be positive work
Q-W = dU
dU = 1.89 kJ - 0.022548 kJ = 1.867452 kJ
dU = 1.867452 kJ increase due to heating, mainly
b)
If no movement exist, the work = 0
so
Q - W = dU
Q = dU
constant volume so use Cv
Q = 1.89 kJ
dU = Q = 1.89 kJ
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