Answer:
<em>181 °C</em>
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Explanation:
Initial pressure = 100 kPa
Initial temperature = 30 °C = 30 + 273 K = 303 K
Final pressure = 1200 kPa
Final temperature = ?
n = 1.2
For a polytropic process, we use the relationship
(/ ) = (/)^γ
where γ = (n-1)/n
γ = (1.2-1)/1.2 = 0.1667
substituting into the equation, we have
(/303) = (1200/100)^0.1667
/303 = 12^0.1667
/303 = 1.513
= 300 x 1.513 = 453.9 K
==> 453.9 - 273 = 180.9 ≅ <em>181 °C</em>
a bricklayer, a bricklayer builds houses repairs walls and chimneys etc.
Answer:
58.44 g/mol The Molarity of this concentration is 0.154 molar
Explanation:
the molar mass of NaCl is 58.44 g/mol,
0.9 % is the same thing as 0.9g of NaCl , so this means that 100 ml's of physiological saline contains 0.9 g of NaCl. One liter of physiological saline must contain 9 g of NaCl. We can determine the molarity of a physiological saline solution by dividing 9 g by 58 g... since we have 9 g of NaCl in a liter of physiological saline, but we have 58 grams of NaCl in a mole of NaCl. When we divide 9 g by 58 g, we find that physiological saline contains 0.154 moles of NaCl per liter. That means that physiological saline (0.9% NaCl) has a molarity of 0.154 molar. We can either express this as 0.154 M or 154 millimolar (154 mM).
We need to define the variables,
So,
Therefore, the probability that the repair time is more than 4 horus can be calculate as,
The probability that the repair time is more than 4 hours is 0.136
b) The probability that repair time is at least 12 hours given that the repair time is more than 7 hoirs is calculated as,
The probability that repair time is at least 12 hours given that the repair time is more than 7 hours is 0.63
Answer:
final volume V2 = 0.71136 m³
work done in process W = -291.24 kJ
heat transfer Q = 164 kJ
Explanation:
given data
mass = 1.5 kg
pressure p1 = 200 kPa
temperature t1 = 150°C
final pressure p2 = 600 kPa
final temperature t2 = 350°C
solution
we will use here superheated water table that is
for pressure 200 kPa and 150°C temperature
v1 = 0.95964 m³/kg
u1 = 2576.87 kJ/kg
and
for pressure 600 kPa and 350°C temperature
v2 = 0.47424 m³/kg
u2 = 2881.12 kJ/kg
so v1 is express as
V1 = v1 × m ............................1
V1 = 0.95964 × 1.5
V1 = 1.43946 m³
and
V2 = v2 × m ............................2
V2 = 0.47424 × 1.5
final volume V2 = 0.71136 m³
and
W = P(avg) × dV .............................3
P(avg) = = = 400 × 10³
put here value
W = 400 × 10³ × (0.71136 - 1.43946 )
work done in process W = -291.24 kJ
and
heat transfer is
Q = m × (u2 - u1) + W .............................4
Q = 1.5 × (2881.12 - 2576.87) + 292.24
heat transfer Q = 164 kJ