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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 kJ |
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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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