Answer:
-6.326 KJ/K
Explanation:
A) the entropy change is defined as:

In an isobaric process heat (Q) is defined as:

Replacing in the equation for entropy
m is the mass and Cp is the specific heat of R134a. We can considerer these values as constants so the expression for entropy would be:
Solving the integral we get the expression to estimate the entropy change in the system

The mass is 5.25 Kg and Cp for R134a vapor can be consulted in tables, this value is 
We can get the temperature at the beginning knowing that is saturated vapor at 500 KPa. Consulting the thermodynamic tables, we get that temperature of saturation at this pressure is: 288.86 K
The temperature in the final state we can get it from the heat expression, since we know how much heat was lost in the process (-976.71 kJ). By convention when heat is released by the system a negative sign is used to express it.

With
clearing for T2 we get:

Now we can estimate the entropy change in the system

The entropy change in the system is negative because we are going from a state with a lot of disorder (high temperature) to one more organize (less temperature. This was done increasing the entropy of the surroundings.
b) see picture.
Answer:
The modulus of resilience is 166.67 MPa
Explanation:
Modulus of resilience is given by yield strength ÷ strain
Yield strength = 500 MPa
Strain = 0.003
Modulus of resilience = 500 MPa ÷ 0.003 = 166.67 MPa
Answer:crane and engine I guess
Explanation:
The heat transferred to and the work produced by the steam during this process is 13781.618 kJ/kg
<h3>
How to calcultae the heat?</h3>
The Net Change in Enthalpy will be:
= m ( h2 - h1 ) = 11.216 ( 1755.405 - 566.78 ) = 13331.618 kJ/kg
Work Done (Area Under PV curve) = 1/2 x (P1 + P2) x ( V1 - V2)
= 1/2 x ( 75 + 225) x (5 - 2)
W = 450 KJ
From the First Law of Thermodynamics, Q = U + W
So, Heat Transfer = Change in Internal Energy + Work Done
= 13331.618 + 450
Q = 13781.618 kJ/kg
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Answer:
The result might require 9 bits to store