Answer:
A) T1 = 269.63 K
T2 = 192.59 K
B) W = -320 KJ
Explanation:
We are given;
Initial volume: V1 = 7 m³
Final Volume; V2 = 5 m³
Constant Pressure; P = 160 KPa
Mass; m = 2 kg
To find the initial and final temperatures, we will use the ideal gas formula;
T = PV/mR
Where R is gas constant of helium = R = 2.0769 kPa.m/kg
Thus;
Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K
Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K
B) world one is given by the formula;
W = P(V2 - V1)
W = 160(5 - 7)
W = -320 KJ
Answer:
45200J
Explanation:
Given parameters:
Heat of vaporization of water = 2260J/g
Mass of steam = 20g
Temperature = 100°C
Unknown:
Energy released during the condensation = ?
Solution:
This change is a phase change and there is no change in temperature
To find the amount of heat released;
H = mL
m is the mass
L is the latent heat of vaporization
Insert the parameters and solve;
H = 20g x 2260J/g
H = 45200J
