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m ΔT
ΔT = Q
c m
t2 - t1 = Q ⇔ t2 =( Q ) + t1
c m (c m)
= ( 42000 ) + 25
(4200 ×0,2)
= 50 + 25 = 75 C.
m= 0,2 kg
t(awal)= 25⁰C
Q= 42.000 j
c= 4.200 j/kg⁰C
Ditanya :
t(akhir) ?
Jawab :
Q= m × c × Δt
Q= m × c × (t(akhir) - t(awal)
42.000 = 0,2 × 4.200 × t(akhir) - 25
42.000 = 840 × t(akhir) - 25
42.000/840= t(akhir) - 25
50= t(ahir) - 25
50 + 20= t(akhir)
t(akhir)= 50 + 25
t(akhir)= 75
Jadi, suhu akhir alumunium tersebut adalah 75⁰C.