Kalor sebanyak 84 kj ditambahkan pada 500 g air yang bersuhu 20 'C. Menjadi berapakah suhu air itu? Kalor jenis air 4.200 J/(Kg.K)

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Q = 84 kJ = 84000 J , m = 500 gr = 0,5 kg. || Q= m.c.delta T || 84000= 0,5×4.200×delta T || 84000 = 2100 delta T || delta T = 40 drajat. || delta T = Takhir - Tawal || T akhir= 60 drajat.
Q = 84 kJ = 84000 J , m = 500 gr = 0,5 kg. || Q= m.c.delta T || 84000= 0,5×4.200×delta T || 84000 = 2100 delta T || delta T = 40 drajat. || delta T = Takhir - Tawal || T akhir= 60 drajat.

Jawabanmu

2014-02-04T13:05:47+07:00
Q = 84 kj = 84000
m = 500 g = 0.5 kg
 t_{1}  = 20 C
C = 4200J/(Kg.K)
 t_{2} = ??

Δt =  \frac{Q}{m*c}
( t_{2}  t_{1}  ) =  \frac{8400}{0.5*4200}  
( t_{2} - 20) = 40 
 t_{2} = 40+20
 t_{2} =  60^{o} C

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