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m = 250 gr
c = 1 kal/gr derajat C
delta T = 100 - 20 = 80 derajat T
m. c. (delta T)
250 . 1 . 80
20.000 kal
kalor = 20.000 kal = 20 kkal
m= 250 gr
T1= 20°c
T2=100°c
delta T = 100- 20 = 80°c
ditanya.= berapa kalor yang diperlukan
dijawab = m.c.delta t
250.1kal/gr.80
250.80
20000 kal/gr
jadi, kalor yang dibutuhkan adalah 20000 kal/gr°C