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0,4 kal/kg°C = 0,0004 kal/g°C
m(logam).c(logam).ΔT(logam) = m(air).c(air).ΔT(air)
100. 0,0004. (80 - x) = 200. 1. (x - 20)
0,0004 (80 - x) = 2 (x -20)
0,032 - 0,0004x = 2x - 40
40,032 = 2,0004x
x = 20,00199...
jadi sekitar ≈ 20,002°C
maaf jika terdapat kesalahan menghitung, semoga bermanfaat :)