Answer:
[tex]\boxed {\boxed {\sf 133,500 \ Joules}}[/tex]
Explanation:
Since we are given the specific heat, mass, and change in temperature, we should use this formula for heat energy.
[tex]q=mc \Delta T[/tex]
The mass is 2 kilograms. Iron's specific heat is 445 Joules per kilogram degree Celsius. The temperature increases by 150 degrees Celsius.
[tex]m= 2 \ kg \\c= 445 \ J/kg \textdegree C\\\Delta T= 150 \textdegree C[/tex]
Substitute the values into the formula.
[tex]q= (2 \ kg)( 445 \ J/kg \textdegree C)(150 \textdegree C)[/tex]
Multiply the first two numbers. The kilograms will cancel.
[tex]q= 890 \ J/\textdegree C (150 \textdegree C)[/tex]
Multiply again. This time the degrees Celsius cancel.
[tex]q= 133, 500 \ J[/tex]
The pan absorbs 133,500 Joules of heat energy.