CP450=1.435+2.19×10−3×450=2.4205 kJ/kg KCP_{450} = 1.435 + 2.19 \times 10^{-3} \times 450 = 2.4205 \;kJ/kg\;KCP450=1.435+2.19×10−3×450=2.4205kJ/kgK
CP550=1.435+2.19×10−3×550=2.6395 kJ/kg KCP_{550} = 1.435 + 2.19 \times 10^{-3} \times 550 = 2.6395 \;kJ/kg\;KCP550=1.435+2.19×10−3×550=2.6395kJ/kgK
∆CP=2.6395–2.4205=0.219 kJ/kg K∆CP = 2.6395 – 2.4205 = 0.219 \;kJ/kg\;K∆CP=2.6395–2.4205=0.219kJ/kgK
The heat capacity was changed by 0.219 units, so we need to supply 0.219 kJ of heat to the heater per kg of the liquid heated.
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