Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.850 kg of water decreased from 109 °c to 48.0 °c.

-2.12 MJ

Explanation:

There are three heat transfers involved.

Heat required = heat to cool vapour + heat to condense vapour to water + heat to cool water

q = q₁ + q₂ + q₃

Step 1: Calculate q₁

q₁ = mC₁ΔT₁

m = 0.850 kg =850 g

C₁ = 1.996 J·°C⁻¹g⁻¹

ΔT₁ = T_f – T_i  = 100 °C – 109 °C = -9 °C

q₁ = 850 × 1.996 × (-9)  

q₁= -15 270 J

The negative sign shows that heat is removed from the system.

Step 2. Calculate q₂

q₂ = mΔH

ΔH = -2260 J·g⁻¹

q₂ = 850 × (-2260)

q₂ = -1 921 000 J

Step 3: Calculate q₃

q₃ = mC₃ΔT₃

C₃ = 4.179 J·°C⁻¹g⁻¹

ΔT₃ = T_f – T_i  = 48.0 °C – 100 °C = -52 °C

q₃ = 850 × 4.179 × (-52) = -184 700 J

Step 4. Calculate the total energy involved

q = q₁ + q₂ + q₃  

q = -15 270 – 1 921 000 – 184 700

q = -2 120 000 J = -2.12 MJ

The energy change is -2.12 MJ.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .