Question

Panci aluminium 0.5 kg suhu awal 15°C diberi kalor 22.500J. C= 90J/kg°C. Berapa suhu akhirnya?

Answer 1

Dik:
m= 0,5 kg
Q= 22500 J
C= 90 J/kg°C
t₁= 15 °C
Dit:
t₂=...?
Penye:
t₂= Δt+t₁
Δt= Q/(m×c)
Δt= 22500/(0,5×90)
Δt= 22500/45
Δt= 500 °C
t₂= 500+15
t₂= 515 °C 

Answer 2

Dik= 
m= 0,5 kg
t₁= 15°C
Q= 22500 J
C= 90 J/kg°C
dit= t₂
jwb=
Q= m.c.Δt
22500=0,5.90.(t₂-t₁)
22500=45(t₂-t₁)
500= t₂-15
t₂=500+15
t₂=515°C