Answer:
Q = 65 kJ
Step-by-step explanation:
Given that,
Mass, m = 580 g
Initial temperature,
![T_i=25^(\circ) C](https://img.qammunity.org/2021/formulas/chemistry/high-school/gunw40533yhohxd9mu4vvxx2pvyyds7hpf.png)
Final temperature,
![T_f=150^(\circ) C](https://img.qammunity.org/2021/formulas/chemistry/college/uitwpegojmsgo8wz69riis7pw0e688hbtd.png)
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 :
![Q=mc\Delta T\\\\Q=580\ g* 0.897\ J/g^(\circ) C* (150-25)^(\circ) C\\\\Q=65032.5\ J](https://img.qammunity.org/2021/formulas/chemistry/college/mjwlmex3e53npo8lkvgnl9rws3dgu7v4m8.png)
or
Q = 65.03 kJ
or
Q = 65 kJ
Hence, the heat added to the pan is 65 kJ.