Answer: The temperature of the aluminium pan was decreased by the 117.25 Kelvins.
Explanation:
Energy given by the aluminium pan =Q= -5400 J
Negative sign indicates that energy is released by the aluminium pan.
Mass of the pan=m = 50 g = 0.05 kg(1000g = 1kg)
Change in temperature =[tex]\Delta T[/tex]
Specif heat of he aluminium = c = 921.096J/kg K
[tex]Q=mc\Delta T[/tex]
[tex]-5400 J=0.05 kg\times 921.096J/kg K\times \Delta T[/tex]
[tex]\Delta T=-117.25 K[/tex]
The negative value of change is temperature means that temperature of the aluminium pan was decreased by the 117.25 Kelvins.
