Question

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 Value Units Melting point Boiling point с 100.0 6.01 A Hus kJ/mol 40.67 ΔΗνωρ Co(s) colo kJ/mol J/mol. °C 37.1 75.3 J/mol." 336 J/mol. "C

Answer 1

Answer:

Step 1: Explanation

When the temperature of n moles of a substance, having specific heat at constant pressure is Cp is changed by ΔT, then the amount of heat transferred by the body is given by

Q = nCpΔT

Step 2: calculation of moles (n)

Given:

Mass of water = 1.25 kg = (1.25 kg  × 1000 g/ 1 kg ) = 1250 g and Molar mass of water = 18.01528 g/mol

So, Moles = Mass / Molar Mass = 1250 g / 18.01528 g/mol = 69.3855 mol

Initial temperature ( T₁ ) = 115 °C

Final temperature ( T₂ ) = 42.5 °C

Step 3: Calculate the energy change for steam from 115 °C to 100 °C

Q₁ = nCp(g)ΔT = 69.3855 mol × 33.6 J / mol. °C ×  (100-115) °C = -34977.6 J

[ Energy released in kJ = ( -34977.6 J × 1 kJ / 1000 J ) = -34.97 kJ ]

Step 4 : Calculate the energy change to convert gaseous water to liquid at 100 °C

Q₂ = -nΔHvap = -69.3855 mol × 40.67 kJ / mol = -2821.91 kJ [ here negative sign due to phase change ]

Step 5: Calculate the energy change for liquid water from 100 °C to 42.5 °C

Q₃ = nCp(l)ΔT = 69.3855 mol × 75.3 / mol. °C ×  (42.5-100) °C = -300422.0584 J

[ Energy released in kJ = ( -300422.0584 J × 1 kJ / 1000 J ) = -300.4220584 kJ ]

Thus, we the total amount of energy change for 1.25 kg of water to go from 115 °C to 42.5°C

Q = Q₁+Q₂+Q₃ = ( -34.97 kJ ) + ( -2821.91 kJ ) +( -300.4220584 kJ ) = -3157.3 kJ