sebuah logam bermassa 7.000 gram dipanaskan hingga bersuhu 100 c kemudian dimasukkan kedalam air bermassa 2.000 gram yang bersuhu 25 c. kalor jenis logam 800j/k

m1 = 2000 gram = 2 kg
m2 = 7000 gram = 7 kg
c1 = 4200 J/kg°C
c2 = 800 J/kg°C
T1 = 25°C
T2 = 100°C
ditanya Tc = ......?
jawab
m1.c1 (Tc - T1) = m2.c1 (T2 - Tc)
2 × 4200 (Tc - 25) = 7 × 800 (100 - Tc)
8400(Tc - 25) = 5600 (100 - Tc) --> masing2 ruas dibagi 400

21 (Tc - 25) = 14(100 - Tc)
21Tc - 525 = 1400 - 14Tc
35Tc = 1925
Tc = 55°C

jawaban C

Suhu dan Kalor.

Q lepas = Q terima
m₁c₁(T₁ - Tc) = m₂c₂(Tc - T₂)
m₁c₁T₁ - m₁c₁Tc = m₂c₂Tc - m₂c₂T₂
m₁c₁T₁ + m₂c₂T₂ = (m₂c₂ + m₁c₁)Tc
Tc = (m₁c₁T₁ + m₂c₂T₂) / (m₁c₁ + m₂c₂)
     = [7(800)(100) + 2(4200)(25)] / [7(800) + 2(4200)]
     = 55°C