Question

Sepotong logam dengan kalor jenis 0.2 kal/gr derajat celcius bermassa 100gram bersuhu 30 derajat celcius dimasukkan pada bejana berisi air yang bersuhu 90 derajat celcius bermassa 200gram. jika kalor jenis air adalah 1 kal/gr derajat celcius dan pengaruh bejana diabaikan tentuka n suhu akhir logam

Answer 1

Diket:
c logam = 0,2 kal/g'c
m logam = 100 g
t2 = 30'c (suhu logam)
t1 = 90'c (suhu air)
m air = 200 g
c air = 1 kal/g'c
Ditanya suhu akhir logam...?
Qlepas = Qterima
mair . c air (t1 - tc) = m logam . c logam (tc - t1)
200 . 1 . (90 - tc) = 100 . 0,2 (tc - 30)
200 (90-tc) = 20 (tc - 30)
18000 - 200tc = 20tc - 600
200tc+20tc = 18000 + 600
220tc = 18600
tc = 84,54'c
suhu logam 84,54'c

Answer 2

dik . c logam = 0,2 kal/g'c
       m logam = 100 g
       t2 = 30'c (suhu logam)
       t1 = 90'c (suhu air)
      m air = 200 g
      c air = 1 kal / g ' c
   
Ditanya suhu akhir logam...?

Qlepas = Qterima
mair . c air (t1 - tc) = m logam . c logam (tc - t1)
200 . 1 . (90 - tc) = 100 . 0,2 (tc - 30)
200 (90-tc) = 20 (tc - 30)
18000 - 200tc = 20tc - 600
200tc+20tc = 18000 + 600
220tc = 18600
tc = 84,54'c

Suhu logam 84,54’c