If 24500 j is applied to 125g of water at 35 c, what will the final temperature of the water be?

The  final  temperature  of  water  is calculated  using the below  equation

Q= MC delta  T
Q= heat  energy (24500  j)
m= mass( 125 g)
c= specific   heat capacity (  4.18  j/g/c)
delta T = change  in temperature ( final temperature -initial  temperature)=(T-35)

final  temperature  is therefore=24500 j =125 g x 4.18 g/j/c  x (T- 35)
24500 j=522.5 (T-35 c)
24500 j = 522.5T -18287.5

like  terms  together  24500 + 18287.5 =522.5 T
42787.5j =522.5 T
divide both  side  by  522.5 T

T=81.9 c

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