LiCl(s) ΔH = –36.9 kJ 25.0 g LiCl 125 mL water 25.0℃

LiCl(s) ΔH = –36.9 kJ 25.0 g LiCl 125 mL water 25.0℃

 

 

LiCl(s) + H2O(l) → Li^+(aq) + Cl^-(aq) ... ΔH = –36.9 kJ

25.0 g LiCl과 물 125 mL를 압착했을 때 최종 온도는 몇 ℃인가?

초기 농도는 25.0℃이고, 용액의 비열은 4.18 J/g•℃이며,

핫팩과 주위 간의 열 전달은 없다고 가정한다.

 

 

What is the final temperature in a squeezed hot pack

that contains 25.0 of LiCl dissolved in 125 mL of water?

Assume a specific heat of 4.18 J/g•℃ for the solution,

an initial temperature of 25.0℃,

and no heat transfer between the hot pack and the environment.

 

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LiCl의 몰질량 = 42.39 g/mol 이므로,

25.0 g / (42.39 g/mol) = 0.58976 mol LiCl

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

0.58976 mol × (36900 J/mol) = 21762 J

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

Δt = q / C m

= 21762 / [(4.18) (25.0+125)]

= 34.7℃

 

 

 

25.0 + 34.7 = 59.7℃

 

 

 

답: 59.7℃

 

 

 

 

[키워드] LiCl 핫팩 기준문서, LiCl hot pack dic, LiCl(s) ΔH = -36.9 kJ 25.0 g LiCl 125 mL water 4.18 J/g•℃ 25.0℃