0.650 Kg of water = 650 gm water = 650 /18 = 36.11 mole.
change in energy = m1 * s1 * dT1 (from 125 oC to 100 oC) + m1 * Lv (for latent heat) + m1 * s2 * dT2 (from 100 oC to 32.0 oC)
change in energy = 36.11 * 33.6 *( 125 - 100) + 36.11 * (40.67*1000) + 36.11 * 75.3 * (100 - 32)
change in energy = 30332.4 + 1468593.7 + 184897.6
change in energy = 1683823.7 J = 1683.8 KJ
887 02/28/18 See Periodic Table See Hint Based on the thermodynamic properties provided for water, determine...
w See Periodic Table See Hint Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.850 kg of water decreased from 125 °C to 37.0 °C. Units Property Melting point Boiling point Value 0 ΔΗ. kJ/mol kJ/mol A Hvap Co (5) co ( 100.0 6.01 40.67 37.1 75.3 33.6 | J/mol. C J/mol. °C J/mol. C se) 587668 kJ > 1st attempt
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Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.25 kg of water decreased from 115 °C to 42.5 °C. Property Melting point Units °C Boiling point °C ΔHus kJ/mol Value 0 100.0 6.01 40.67 37.1 75.3 33.6 kJ/mol J/mol. "C ΔHap (s) 40( J/mol. °C J/mol. °C Co (8)
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