Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.750 kg of water decreased from 119 °C to 16.0 °C. Property Value Units Melting point 0 °C Boiling point 100.0 °C ΔHfus 6.01 kJ/mol ΔHvap 40.67 kJ/mol cp (s) 37.1 J/mol · °C cp (l) 75.3 J/mol · °C cp (g) 33.6 J/mol · °C

Answer 1

Heat change to decrease temperature of water (gas) from 119⁰ to 100^o c.

Moles of water = (mass/ molar mass)

= ( 0.750*1000) / 18 = 41.67

q₁ = moles of water × specific heat of water (gas) ×

= 41.67 × 33.6 × ( 100 - 119) J

= - 26600 J.

= - 26.60 KJ

Heat change to convert 100⁰ c water (gas) to 100^o water (liquid)

q₂ = - moles of water × heat of vaporization

= - 41.67 × 40.67

= - 1694.72 KJ .

Heat change to convert water (liquid) from 100^oc to 16⁰c

q₃ = moles of water × specific heat ×

= 41.67 × 75.3 × (16 - 100)

= - 263571 J.

= - 263.57 KJ.

Total heat ( energy ) change to convert 119^o to 16⁰ c

= q₁+ q₂ +q₃

= ( -26.60 - 1694.72 - 263.57 ) KJ .

= - 1984.89 KJ .