Apakah jawaban dari pertanyaan di bawah ini

m = 3 kg es
c = 2.100 J/kg°C = kalor jenis
L = 336.000J/kg = kalor lebur
Delta T = 20 - (10) = 30 (A - C) (perubahan suhu)

Kalor = Q1 = m . c. Delta
= 3 . 2100 .10 = 6300 x 10 = 63000 J

Kalor = Q2 = m . L
= 3 x 336000
= 1.008.000 J

Kalor total = Q1 + Q2 = 63000 + 1008000 = 1071000 j
= 1071 kj