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Answer:
P₂ ≅ 100 atm (1 sig. fig. based on the given value of P₁ = 90 atm)
Explanation:
Given:
P₁ = 90 atm P₂ = ?
V₁ = 18 Liters(L) L₂ = 12 Liters(L)
=> decrease volume => increase pressure
=> volume ratio that will increase 90 atm is (18L/12L)
T₁ = 272 Kelvin(K) T₂ = 274 Kelvin(K)
=> increase temperature => increase pressure
=> temperature ratio that will increase 90 atm is (274K/272K)
n₁ = moles = constant n₂ = n₁ = constant
P₂ = 90 atm x (18L/12L) x (274K/272K) = 135.9926471 atm (calculator)
By rule of sig. figs., the final answer should be rounded to an accuracy equal to the 'measured' data value having the least number of sig. figs. This means P₂ ≅ 100 atm based on the given value of P₁ = 90 atm.
Explanation:
Apply the mass of balance as follows.
Rate of accumulation of water within the tank = rate of mass of water entering the tank - rate of mass of water releasing from the tank



[/tex]\frac{dh}{dt} + \frac{0.01}{0.01}h[/tex] = 

+ h = 1
= 1 - h
= dt
= t + C
Given at t = 0 and V = 0
= 0
or, h = 0
-ln(1 - h) = t + C
Initial condition is -ln(1) = 0 + C
C = 0
So, -ln(1 - h) = t
or, t =
........... (1)
(a) Using equation (1) calculate time to fill the tank up to 0.6 meter from the bottom as follows.
t =
t =
= 
= 0.916 seconds
(b) As maximum height of water level in the tank is achieved at steady state that is, t =
.
1 - h = exp (-t)
1 - h = 0
h = 1
Hence, we can conclude that the tank cannot be filled up to 2 meters as maximum height achieved is 1 meter.
Heat energy is needed for evaporation to happen.
<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>