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M=1,5 kg
t1= 20 °C
Q= 378 joule
C= 4200 J/kg°C
Ditanya=t2.....?
Jawab=
t2=Δt+t1
Δt= Q/(MxC)
Δt= 378/(1,5 x 4200)
Δt= 378/6300
Δt= 0,06°C
t2= ∆t+t1
t2= 0,06 + 20
t2= 20,06°C
(Semoga membantu,jadikan jawaban yang terbaik ya,terima kasih