Answer:
2400°C
Explanation:
Volume of tank = 0.4 m^3
steam pressure = 6 Mpa
Steam temperature = 40°C
Initial pressure of tank = 0.1 MPa
Final pressure of Tank = 6 Mpa
Calculate the temperature in the tank when the Pressure in Tank = 6Mpa
since the volume of the Tank is constant = 0.4 m^3
we will apply Gay-Lussac's Law
= [tex]\frac{T1}{P1} = \frac{T2}{P2}[/tex] ------ ( 1 )
T1 = 40°c
P1 = 0.1 MPa
P2 = 6 Mpa
T2 = ?
From equation 1 above
T2 = ( T1 * P2 ) / P1
= ( 40 * 6 ) / 0.1
= 2400°C
