Question

887 02/28/18 See Periodic Table See Hint Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.650 kg of water decreased from 125 ℃ to 320 °C、 Value Property Melting point Boiling point Units oc 1000 6.01 4067 37.1 75.3 33.6 Wmol C 23 oF 28 qUESTIONS COMPLETE 28-28 command option
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Answer #1

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

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