Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid water from room temperature (20 °C) to its normal boiling point (100 °C), while still remaining a liquid.(a) 104.4 kJ(b) 100.4 kJ(c) 140.4 kJ(d) 104.5 kJThis question was addressed to me in final exam.Asked question is from Enthalpy Change in Non-Reactive Processes in portion Energy Balance of Bioprocess Engineering

Calculate how much energy (kJ) is required to heat a bottle containing

1.	Calculate how much energy (kJ) is required to heat a bottle containing 300 mL of liquid water from room temperature (20 °C) to its normal boiling point (100 °C), while still remaining a liquid.(a) 104.4 kJ(b) 100.4 kJ(c) 140.4 kJ(d) 104.5 kJThis question was addressed to me in final exam.Asked question is from Enthalpy Change in Non-Reactive Processes in portion Energy Balance of Bioprocess Engineering
Answer» The CORRECT answer is (b) 100.4 KJ Explanation: Enthalpy (“Water”, 20, 000,”l”) = 6.032 kJ/mol \((∫_{20}^{100} C_{pH_2O} dT)\) = \( (\frac{6.032 kJ}{mol}) (\frac{1g}{ml}) (\frac{mol}{18.02 G}) (\frac{300 ml}{1})\) = 100.4 kJ

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