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

User Ancy
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1 Answer

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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


User OJVM
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