491.8 J heat 23.0℃ 5.00 g water final temp.

491.8 J heat 23.0℃ 5.00 g water final temp.

 

 

Assume that 491.8 J of heat is added to 5.00 g of water originally at 23.0℃.

What would be the final temperature of the water?

(Specific heat capacity of water = 4.184 J/g•℃.)

 

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q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

 

 

Δt = q / C m

= (491.8 J) / [(4.184 J/g•℃) (5.00 g)]

= 23.5℃

 

 

 

23.0 + 23.5 = 46.5℃

---> final temperature of the water