Both portions of the rod ABC are made of an aluminum for whichE = 70 GPa. Knowing that the magnitude of P is 4 kN, determine(a)
1 answer:
You might be interested in
Answer:
Given:
P₁ = 500 kPa
T₁ = 860 K
P₂= 100 kPa
T₂ = 460 K
Let's take entropy properties of T1 and T2 from ideal properties of air,
at T = 860K, s(T₁) = 2.79783 kJ/kg.K
at T = 460K, s(T₂) = 2.13407 kJ/kg.K
using entropy balance equation:
= - 0.2018 kJ/kg. K
In this case the entropy is negative, which means the value of exit temperature is not correct, beacause entropy should always be positive(>0).
Answer:
5000J
Explanation:
Given in the question that
Heat added to the coffee cup is, ΔS = 20 J/K
The temperature of the coffee, T = 250 K
Now, using the formula for the entropy change
...........(1)
Where,
ΔS is the entropy change
ΔH is the enthalpy change
T is the temperature of the system
substituting the values in the equation (1)
we get
ΔH=250×20
ΔH=5000 J