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= 0,1 kg
ΔT1 = 0 - ( -10 )
= 10°C
ΔT2 = 10 - 0
= 10°C
c1 = 0,5 kal/g°C
= 0,5 kal / 0,001 kg°C
= 500 Kal/kg°C
c2 = 1 Kal/g°C
= 1000 Kal/kg°C
L = 80 Kal/kg
= 8 x 10^4 kal/kg
Q1 = m . c1 . ΔT1
= 0,1 . 500 . 10
= 500 Kal
Q2 = m . L
= 0,1 . 80 000
= 8 000 Kal
Q3 = m . c2 . ΔT2
= 0,1 . 1000 . 10
= 1000 Kal
Qtotal ?
Qt = Q1 + Q2 + Q3
= 500 + 8000 + 1000
= 9500 Kal ( D )