50.0 g metal 100.0℃ 24.8℃ 50.0 g water 51.2℃
Suppose 50.0 g of water in the Styrofoam cup has an in initial temperature of 24.8℃. When a 50.0 g piece of metal at a temperature of 100.0℃ is added, the temperature of the water and metal rise to final temperature of 51.2℃. What is the specific heat of this metal object? (Specific heat of water is 4.184 J/g•℃)
열용량과 비열
q = C m Δt
( 참고 https://ywpop.tistory.com/2897 )
금속의 비열 = x 라 두면,
고온 물질이 잃은 열(–q) = 저온 물질이 얻은 열(+q)
–[(x) (50.0) (51.2 – 100.0)] = +[(4.184) (50.0) (51.2 – 24.8)]
2440x = 5522.88
x = 5522.88 / 2440
= 2.2635 J/g•℃
---> 금속의 비열
답: 2.26 J/g•℃
[ 관련 글 https://ywpop.tistory.com/23649 ]
metal sample 110.0 g 92.0℃ heat 21.0℃ water 75.0 g 24.2℃
[키워드] 고온 물질이 잃은 열 기준
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