ΔH = +25.7 kJ 50.0 g NH4NO3 water 125 mL 4.18 J 25.0℃

ΔH = +25.7 kJ 50.0 g NH4NO3 water 125 mL 4.18 J 25.0℃

 

 

Instant cold packs are effective

because the dissolution of solid NH4NO3 is endothermic:

NH4NO3(s) + H2O(l) → NH4NO3(aq) ... ΔH = +25.7 kJ/mol

 

What is the final temperature in a squeezed cold pack

that contains 50.0 g of NH4NO3 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 cold pack and the environment.

 

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[ 블로그스팟 https://ywpop.blogspot.com/2023/10/h-257-kj-500-g-nh4no3-water-125-ml-418.html ]

 

 

 

NH4NO3의 몰질량 = 80.04 g/mol

50.0 g / (80.04 g/mol) = 0.6247 mol NH4NO3

( 참고 https://ywpop.tistory.com/7738 )

 

 

 

0.6247 mol × (25.7 kJ/mol) = 16.1 kJ = 16100 J

---> 이만큼의 열을 팩이 빼앗는다.

 

 

 

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

 

Δt = q / C m

= 16100 J / [(4.18 J/g•℃) (50.0 + 125 g)]

= 22.0℃

---> 흡열이므로, 이만큼 온도가 낮아질 것이다.

 

 

 

25.0 – 22.0 = 3.0℃

 

 

 

답: 3.0℃

 

 

 

 

[공식] Δt 계산

= [ NH4NO3의 질량(g) / 80.04 × ΔH(kJ/mol) × 1000 ]

/ [ 용액의 비열 × (NH4NO3의 질량(g) + 물의 질량(g)) ]

= [50.0 / 80.04 × 25.7 × 1000] / [4.18 × (50.0 + 125)]

= 21.9473℃

 

 

 

[키워드] 냉찜질 팩 기준, 냉찜질 기준, NH4NO3 기준, 흡열반응 기준

 

[FAQ] [①23/10/24]