Question

Tolong jawab pakai caranya

Answer 1

Qa-b = m . c . delta t
= 3 . 2100 . 10
= 63.000 j
Qb-c = m . l
= 3 . 336.000
= 1.008.000 j
Qc-d = m. c. delta t
=3. 4200 .80
= 1.008.000 j
Qa-d = 63.000 + 1.008.000 + 1.008.00 = 2.079.000 j = 2.079 kj

Answer 2

Dik:
kalor jenis es = 2100 J/kg°C
kalor lebur es = 33600 J/kg
Kalor jenis air = 4200 J/Kg °C

Dit: q
jwb:
q= MxCxdelta T + M x L + M x C x delta T
= 3kg x 2100 J/kg°C × 10°C + 3 kg x 336000 J/kg + 3 kg x 4200 J/kg° x 80°C
=2079000 J / 2079 kJ