Respuesta :
Answer:
the activation energy Ea = 179.176 kJ/mol
it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
Explanation:
From the given information
[tex]T_1 = 100^0 C = 100+273 = 373 \ K \\ \\ T_2 = 113^0 C = 113 + 273 = 386 \ K[/tex]
[tex]R_1 = \dfrac{1}{7}[/tex]
[tex]R_2 = \dfrac{1}{49}[/tex]
Thus; [tex]\dfrac{R_2}{R_1} = 7[/tex]
Because at 113.0°C; the rate is 7 time higher than at 100°C
Hence:
[tex]In (7) = \dfrac{Ea}{8.314}( \dfrac{1}{373}- \dfrac{1}{386})[/tex]
1.9459 = [tex]\dfrac{Ea}{8.314}* 9.0292 *10^{-5}[/tex]
[tex]1.9459*8.314 = Ea * 9.0292*10^{-5}[/tex]
[tex]16.1782126= Ea * 9.0292*10^{-5}[/tex]
[tex]Ea = \dfrac{16.1782126}{ 9.0292*10^{-5}}[/tex]
Ea = 179.176 kJ/mol
Thus; the activation energy Ea = 179.176 kJ/mol
b)
here;
[tex]T_2 = 386 \ K \\ \\T_1 = (89.8 + 273)K = 362.8 \ K[/tex]
[tex]In(\dfrac{R_2}{R_1})= \dfrac{Ea}{R}(\dfrac{1}{T_1}- \dfrac{1}{T_2})[/tex]
[tex]In(\dfrac{R_2}{R_1})= \dfrac{179.176}{8.314}(\dfrac{1}{362.8}- \dfrac{1}{386})[/tex]
[tex]In (\dfrac{R_2}{R_1}) = 0.00357[/tex]
[tex]\dfrac{R_2}{R_1}= e^{0.00357}[/tex]
[tex]\dfrac{R_2}{R_1}= 1.0035[/tex]
where ;
[tex]R_2 = \dfrac{1}7{}[/tex]
[tex]R_1 = \dfrac{1}{t}[/tex]
Now;
[tex]\dfrac{t}{7}= 1.0035[/tex]
t = 7.0245 mins
Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
a). The activation energy given by the reactions related to the cooking of food in the pressure cooker would be:
[tex]Ea = 179.176 kJ/mol[/tex]
b). The time duration that is taken by the same food to cook in an open vessel would be:
[tex]7.0245 mins[/tex]
Activation Energy
a). Given that,
Temperature [tex]1[/tex] [tex]= 100[/tex]° C
Temperature [tex]2[/tex] [tex]= 113[/tex]° C
In Kelvin,
Temperature [tex]1[/tex] [tex]= 100 + 273[/tex]
[tex]= 373 K[/tex]
Temperature [tex]2[/tex] [tex]= 113 + 273[/tex]
[tex]= 386 K[/tex]
[tex]R_{1} = 1/7\\R_{2} = 1/49[/tex]
∵ [tex]R_{2}/R_{1} = 49/7 = 7[/tex]
It is given that at [tex]113[/tex] rate [tex]=[/tex] [tex]7[/tex] × [tex]100[/tex]°C
Therefore,
[tex]Ea/8.314 (1/373 - 1/386) =[/tex] [tex]In(7)[/tex]
so,
[tex]Ea[/tex] [tex]= 16.1782126/(9.0292 * 10^{-5})[/tex]
∵ Activation energy [tex]= 179.176 kJ/mol[/tex]
b). As we know,
[tex]T_{2}[/tex] [tex]= 386 K[/tex]
[tex]T_{1}[/tex] [tex]= (89. 8 + 273)[/tex]
[tex]= 362.8 K[/tex]
by employing the formulae,
[tex]In(\frac{R_{2} }{R_{1} }) = \frac{Ea}{R} (1/T_{1} - 1/T_{2})[/tex]
[tex]In(\frac{R_{2} }{R_{1} }) = 179.176/8.314 (1/362.8 - 1/386)[/tex]
By solving this, we get
[tex]R_{2}/R_{1} = 1.0035[/tex]
Thus,
[tex]R_{2} = 1/7[/tex]
[tex]R_{1} = 1/t[/tex]
∵ t [tex]= 7.0245 min[/tex]
Thus, the time duration would be [tex]7.0245 minutes[/tex].
Learn more about "Boiling Point" here:
brainly.com/question/2153588
