Question

The instruction booklet for your pressure cooker indicates that its highest setting is 12.3 psi . you know that standard atmospheric pressure is 14.7 psi, so the booklet must mean 12.3 psi above atmospheric pressure. at what temperature in degrees celsius will your food cook in this pressure cooker set on "high"

Answer 1

Food in a pressure cooker set on "high" at 12.3 psi above atmospheric pressure, equalling a total of 27 psi, will likely cook at a temperature slightly below 120°C. This is based on the general functioning of pressure cookers, which can reach up to 122°C at around 2 atm pressure.

Step-by-step explanation:

The temperature at which your food will cook in a pressure cooker set on "high" can be determined by understanding how pressure affects the boiling point of water. The instruction booklet states that the highest setting is 12.3 psi above atmospheric pressure. Taking into account that standard atmospheric pressure is 14.7 psi, the total pressure inside the pressure cooker would be 12.3 psi + 14.7 psi = 27 psi (or approximately 1.86 atm, since 1 atm = 14.7 psi).

At higher pressures, water boils at a higher temperature. However, without a specific boiling point to temperature correlation for 1.86 atm, we can use the fact that at standard atmospheric pressure water boils at 100°C and typical pressure cookers can reach temperatures of up to 120°C to 122°C at pressures around 2 atm. Given this information, we can infer that food in this pressure cooker would likely cook at a temperature slightly below 120°C, since 1.86 atm is slightly lower than 2 atm.

While an exact boiling point requires more detailed phase information or empirical data for the substance at that pressure, an estimation can be made based on the typical functioning of pressure cookers and what is known about boiling points at different pressures.

Answer 2

118 C The Clausius-Clapeyron equation is useful in calculating the boiling point of a liquid at various pressures. It is: Tb = 1/(1/T0 - R ln(P/P0)/Hvap) where Tb = Temperature boiling R = Ideal Gas Constant (8.3144598 J/(K*mol) ) P = Pressure of interest Hvap = Heat of vaporization of the liquid T0, P0 = Temperature and pressure at a known point. The temperatures are absolute temperatures. We know that water boils at 100C at 14.7 psi. Yes, it's ugly to be mixing metric and imperial units like that. But since we're only interested in relative pressure differences, it's safe enough. So P0 = 14.7 P = 14.7 + 12.3 = 27 T0 = 100 + 273.15 = 373.15 And for water, the heat of vaporization per mole is 40660 J/mol Let's substitute the known values and calculate. Tb = 1/(1/T0 - R ln(P/P0)/Hvap) Tb = 1/(1/373.15 K - 8.3144598 J/(K*mol) ln(27/14.7)/40660 J/mol) Tb = 1/(0.002679887 1/K - 8.3144598 1/K ln(1.836734694)/40660) Tb = 1/(0.002679887 1/K - 8.3144598 1/K 0.607989372/40660) Tb = 1/(0.002679887 1/K - 5.055103194 1/K /40660) Tb = 1/(0.002679887 1/K - 0.000124326 1/K) Tb = 1/(0.002555561 1/K) Tb = 391.3034763 K Tb = 391.3034763 K - 273.15 Tb = 118.1534763 C Rounding to 3 significant figures gives 118 C