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2 answers:
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Answer:
The pressure inside the cooker is 1.0804 atm.
Explanation:
Boiling occurs when the vapor pressure becomes equal to atmospheric pressure.
<u>For water, At standard conditions (Pressure = 1 atm) boiling occurs at 373.15 K.</u>
So, Standard conditions:
T₁ = 373.15 K
P₁ = 1 atm
Given ,
The water boils at Temperature = 130 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So, the temperature, T₂ = (130 + 273.15) K = 403.15 K
To find pressure inside the cooker (P₂) :
<u>Applying Amontons's Law as:</u>
So,
<u>Thus, The pressure inside the cooker is 1.0804 atm.</u>
Answer:
surface temp of fuel rod = 678.85 K
Explanation:
Given data :
D1 = 25 mm
D2 = 50 mm
T2 = 335 k
T∞ = 300 k
hconv = 0.15 w/m^2.k
ε2 = 0.05
ε1 = 1
Determine energy at Q23
Q23 = Qconv + Qrad
attached below is the detailed solution
Insert given values into equation 1 attached below to obtain the surface temperature of the fuel rod ( T1 )
