Respuesta :
Answer:At [tex]237.57^oC[/tex] food will cook in the pressure cooker set on high.
Explanation:
Standard atmospheric pressure [tex]P_1[/tex]= 14.7 Psi = 1.0001 atm (1 Psi=0.06804 atm)
Standard temperature[tex]T_1[/tex] = 273.15 K
Highest pressure offered by pressure cooker: 12.8 Psi = 0.8709 atm
Pressure ([tex]P_2[/tex]) inside the pressure cooker on the highest setting at temperature[tex]T_2[/tex] :
[tex]P_2[/tex] = 1.0001 atm + 0.8709 atm = 1.8710 atm
According to Gay lussac law :
[tex](pressure)\propto (temperature)[/tex] (at constant Volume)
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
[tex]T_2=\frac{P_2\times T_1}{P_1}=\frac{1.8710 atm\times 273.15 K}{1.0001 atm}=510.73 K= 237.57^oC (T(^oC)=T-273.15 K)[/tex]
At [tex]237.57^oC[/tex] food will cook in the pressure cooker set on high.
Answer: The temperature in the pressure cooker set on high is 237.71°C
Explanation:
STP conditions:
Pressure = 14.7 psi
Temperature = 273 K
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
where,
[tex]P_1\text{ and }T_1[/tex] are the initial pressure and temperature of the gas.
[tex]P_2\text{ and }T_2[/tex] are the final pressure and temperature of the gas.
We are given:
[tex]P_1=14.7psi\\T_1=273K\\P_2=(14.7+12.8)psi=27.5psi\\T_2=?[/tex]
Putting values in above equation, we get:
[tex]\frac{14.7psi}{273K}=\frac{27.5psi}{T_2}\\\\T_2=510.71K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]510.71=T(^oC)+273\\T(^oC)=237.71^oC[/tex]
Hence, the temperature in the pressure cooker set on high is 237.71°C