Answer:
-20.158ft-lb
Explanation:
Check the attached files for the explanation.
Answer:
the third one is incorrect
Explanation:
10 x 10³= 10^1 x 10^3 = 10^4
<span>12-50t=70t, t= 0.1h = 6 minutes.</span>
Hey there!
In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).
Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:
![x=\frac{m_{steam}}{m_{total}}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7Bm_%7Bsteam%7D%7D%7Bm_%7Btotal%7D%7D)
Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:
![x=\frac{2kg}{10kg} =0.20](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B2kg%7D%7B10kg%7D%20%3D0.20)
Now, at this temperature and pressure, the volume of a saturated vapor is 2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:
![v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg](https://tex.z-dn.net/?f=v%3D0.001036m%5E3%2Fkg%2B0.2%282.3593-0.001036%20%29m%5E3%2Fkg%3D0.4727m%5E3%2Fkg)
This means that the volume of the container will be:
![V=10kg*0.4727m^3/kg\\\\V=4.73m^3](https://tex.z-dn.net/?f=V%3D10kg%2A0.4727m%5E3%2Fkg%5C%5C%5C%5CV%3D4.73m%5E3)
Regards!