Jawaban:
V = 200 ml = 200 cm³ = 200 × 10^(-6) m³
m = ρ × V
m = 500 kg/m³ × 200 × 10^(-6) m³
m = 0,1 kg
Q = m × c × ΔT
4.000 J = 0,1 kg × c × (50°C - 30°C)
4.000 J = 0,1 kg × c × 20°C
4.000 J = 0,002 kg°C × c
c = 4.000 J / (0,002 kg°C)
c = 2.000 J/kg°C
Jadi, kalor jenis cairan X adalah 2.000 J/kg°C.
[tex]\colorbox{black}{\blue{ \boxed{ \tt{\color{aqua}{answered \: by \: BranlySuhuPTK}}}}}[/tex]
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Verified answer
Jawaban:
V = 200 ml = 200 cm³ = 200 × 10^(-6) m³
m = ρ × V
m = 500 kg/m³ × 200 × 10^(-6) m³
m = 0,1 kg
Q = m × c × ΔT
4.000 J = 0,1 kg × c × (50°C - 30°C)
4.000 J = 0,1 kg × c × 20°C
4.000 J = 0,002 kg°C × c
c = 4.000 J / (0,002 kg°C)
c = 2.000 J/kg°C
Jadi, kalor jenis cairan X adalah 2.000 J/kg°C.
[tex]\colorbox{black}{\blue{ \boxed{ \tt{\color{aqua}{answered \: by \: BranlySuhuPTK}}}}}[/tex]