Answer:
Q = 65 kJ
Explanation:
Given that,
Mass, m = 580 g
Initial temperature, [tex]T_i=25^{\circ} C[/tex]
Final temperature, [tex]T_f=150^{\circ} C[/tex]
The specific heat capacity for aluminum is 0.897 J/g°C
We need to find the heat added to the aluminium pan so that its heat raised to a temperature from 25°C to 150°C. It is given by :
[tex]Q=mc\Delta T\\\\Q=580\ g\times 0.897\ J/g^{\circ} C\times (150-25)^{\circ} C\\\\Q=65032.5\ J[/tex]
or
Q = 65.03 kJ
or
Q = 65 kJ
Hence, the heat added to the pan is 65 kJ.