mass of metal (m1) = 192g
T1 = 100°C
c1 (shc of metal) = ?
mass of calorimeter (m2) = 0.128kg = 128g
c2 (shc of calorimeter) = 394 J/kg.K = 0.394 J/g.K
mass of water (m3) = 240g
c3 (shc of water) = 4.184 J/g.K
T2 (Temperature of water and calorimeter) = 8.4°C
T3 (Final Temperature) = 21.5°C
heat lost = heat gained
m1c1(T1-T3) = m2c2(T3-T2) + m3c3(T3-T2)
192 × c1 × (100-21.5) = 128 × 0.394 × (21.5-8.4) + 240 × 4.184 × (21.5 - 8.4)
15,072c1 = 660.7 + 13,154.5
15,072c1 = 13815.2
c1 = 13815.2/15,072
c1 = 0.916 J/g.C
According to Dulong and Petit's law,
Atomic mass × specific heat = 3R
Atomic mass × c = 24.942
Atomic mass = 24.942/0.916 = 27.23g
Therefore, the unknown metal is probably Aluminium. Since the atomic mass of Aluminium is 27g.
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