f 325 g of water at 4.2°C absorbs 12.28 kJ, what is the final temperature of the water? The specific heat of water is 4.184 J/g·°C

Specific heat can be defined as the amount of heat required to raise the temperature of 1 gram of a substance by 1 degree celsius.

The specific heat can be expressed as:

Heat = mass [tex]\times[/tex] specific heat [tex]\times[/tex] (Final temperature - Initial temperature)

From the given:

12.28 kJ = 325 g [tex]\times[/tex] 4.184 J/g[tex]\rm ^\circ C[/tex] [tex]\times[/tex] (Final temperature - 4.2 [tex]\rm ^\circ C[/tex])

12.28 [tex]\times[/tex] 1000 J = 1,359.8 (Final temperature - 4.2 [tex]\rm ^\circ C[/tex] )

12,280 = 1,359.8 (Final temperature - 4.2 [tex]\rm ^\circ C[/tex] )

9.03 = Final temperature - 4.2 [tex]\rm ^\circ C[/tex]

Final temperature = 13.23 [tex]\rm ^\circ C[/tex]

The final temperature of the water after the absorption of 12.28 kJ of energy has been 13.23 [tex]\rm ^\circ C[/tex].

For more information about specific heat, refer to the link:

https://brainly.com/question/22238883