Ten kilograms of water enters the heater at 16 0 C and leaves at 123 0 F. Determine
the heat added in KJ.
Q=cm(t2−t1)==4200⋅10⋅(50.6−16.0)==1453 kJ.Q=cm(t_2-t_1)=\\=4200\cdot 10\cdot(50.6-16.0)=\\=1453~\text{kJ}.Q=cm(t2−t1)==4200⋅10⋅(50.6−16.0)==1453 kJ.
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