500 gram es bersuhu -12°C dipanaskan hingga suhunya 10°C. Jika kalor jenis es 0,5 kal/gr.°C, kalor lebur es 80 kal/gr, dan kalor jenis air 1 kal/gr.°C. Tentukan

es -12* ke 0*
Q1 = mct
 = 500 x 0,5 x (0- (-12))
 = 500 x 0,5 x 12
 = 3000

es melebur
Q2 = m L
 = 500 x 80
 = 40000

air 0* ke 10*
Q3 = mct
 = 500 x 1 x (10-0)
 = 500 x 1 x 10
 = 5000

jumlahkan Q1 Q2 dan Q3

Dik :
m = 500 gram
Ta = -12 °C
Tb = 0 °C
Tc = 10 °C
c es = 0,5 kal/gr.°C
L = 80 kal/gr
c air = 1 kal/gr.°C

Dit : Qtotal =...?

Jawab

Qtotal = Q1 + Q2 + Q3

Q1 = m.c es. delta T1
= 500. 0,5. (Tb - Ta)
= 250. (0 - (-12))
= 250. 12
= 3000 kal = 3 kkal

Q2 = m.L
= 500. 80
= 40000 kal = 40 kkal

Q3 = m.c air. delta T2
= 500.1. (Tc - Tb)
= 500. (10-0)
= 500. 10
= 5000 kal = 5 kkal

Q total = Q1 + Q2 + Q3 = 3 + 40 + 5 = 48 kkal