Question

Besar kalor yang diperlukan dalam percobaan K ke M adalah?​

Answer 1

Verified answer

Kalor

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[tex]Kalor = KL + LM[/tex]

[tex]Kalor = (m.c_{es}. ∆t) + (m.L)[/tex]

[tex]Kalor = (0.2 \: kg \times 2100 \times 5) + (0.2 \: kg \times 336000)[/tex]

[tex]Kalor = 2100 + 67200[/tex]

[tex]Kalor = \boxed{69300~J} \: \: \: \: \: opsi \: c[/tex]

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[tex]\begin{array}{lr}\texttt{Selamat Menunaikan Ibadah Puasa}\end{array}[/tex]

[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 30 - 03 - 2023}}[/tex]

Answer 2

KALOR YANG DIPERLUKAN :

= (0,2(2.100(5) + (0,2(336.000)

= (2/10(10.500) + (2/10(336.000)

= (2(10.500)/10) + (2(336.000)/10)

= (21.000/10) + (672.000/10)

= (2.100 + 67.200)

= 69.300 Joule