The water in a pressure cooker boils at a temperature greater than 100 degree C
ID: 537943 • Letter: T
Question
The water in a pressure cooker boils at a temperature greater than 100 degree C because it is under pressure. At this higher temperature, the chemical reactions associated with the cooking of food take place at a greater rate. (a) Some food cooks fully in 6.00 min in a pressure cooker at 120.0 degree C and in 48.0 minutes in an open pot at 100.0 degree C. Calculate the average activation energy for the reactions associated with the cooking of this food. kJ mol^-1 (b) How long will the same food take to cook in an open pot of boiling water at an altitude of 8500 feet, where the boiling point of water is 91.4 degree C? minExplanation / Answer
(a.) By using the equatin
ln (k2 / k1) = (Ea/R) (1/T1 - /T2)
where T1 = 120 C = 120 + 273 = 393 K
T2 = 100 + 273 = 373 K
Rate constant is inversely proportional to time.
Therefore, k1 = 1/6min-1 and k2 = 1/48 min-1
R = 8.314 J/mol K
Substituting all these values in the above equation, we get:
ln(6/48) = Ea/8.314 (1/393 – 1/373)
-2.07944154 = Ea/8.314 *( -0.000136436)
Ea = (-2.07944154 *8.314 )/(-0.000136436)
Ea = 126715.0276 J/mol
Ea = 126.72 kJ/mol
b) At 8500 feet, water boils at 91.4 oC, T1 = 91.4 +273 K (364.4 K) at t1 = ? min
At sea level, food boils at 100 oC, T2 = 373 K and time 48 min
Applying the same equation:
ln (k2 / k1) = (Ea/R) (1/T1 - /T2)
ln (t1/48) = (126715.02 J/mol)/8.314 J/molK [1/364.4 – 1/373]
ln (t1/48) = 15241.16198* [6.3272 x 10-5]
= 0.964338104
t1/48 = e(0.964338104)
t1 = 48 *2.623
t1 = 125.9 min
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