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