Answer:
118ÂșC
Explanation:
For the Clausius-Clapeyron equation, we know that:
[tex]ln\frac{P2}{P1} = -\frac{L}{R} x(\frac{1}{T2} - \frac{1}{T1} Â )[/tex]
P1 is the atmospheric pressure, P1 = 14.7 psi
P2 is the pressure of the pressure cooker, P2 = 11.9 + 14.7 = 26.6 psi
L is the spefic latent heat of water, L = 40,700 J/K
R is the gas constant, R = 8.314 J/mol*K
T1 is the boling point of water at atmospheric pressure, T1 = 100ÂșC = 373 K
And T2 is the temperature of the pressure cooker, so:
ln (26.6/14.7) = -40,700/8.314*(1/T2 - 1/373)
0.5931 = -4895.36*(1/T2 - 0.00268)
(1/T2 - 0.00268) = - 0.000121
1/T2 = 0.002559
T2 = 391 K
T2 = 118ÂșC