Question

5400 J of energy are given off when a 50 g of aluminum pan changes temperature. Does the Temperature of the pan increase or decrease and by how much

Answer 1

Answer: The temperature of the aluminium pan was decreased by the 117.25 Kelvins.

Step-by-step 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 =
`\Delta T`

Specif heat of he aluminium = c = 921.096J/kg K

`Q=mc\Delta T`

`-5400 J=0.05 kg* 921.096J/kg K* \Delta T`

`\Delta T=-117.25 K`

The negative value of change is temperature means that temperature of the aluminium pan was decreased by the 117.25 Kelvins.