Answer:
[tex]278.25^0C[/tex]
Explanation:
In this case, Gay-Lussac's law is used to compute the required temperature, as it relates pressure and temperature, by knowing that the atmospheric standard temperature is 25 °C, as follows:
[tex]P_1T_2=P_2T_1[/tex]
Thus, the initial pressure is known as 14.7 psi and the initial temperature 25 °C, which correspond to the atmospheric conditions. Nonetheless, the pressure into the cooker, should be:
[tex]P_2=12.5psi+14.7psi=27.2psi[/tex]
Since the 12.5 psi are above the atmospheric pressure. In such a way, the temperature inside the cooker turns out:
[tex]T_2=\frac{P_2*T_1}{P_1}=\frac{27.2psi*(25+273.15)K}{14.7psi} =551.4K=278.25^0C[/tex]
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