Question

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.

Answer 1

-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.

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Step 2. Calculate q₂

q₂ = mΔH

ΔH = -2260 J·g⁻¹

q₂ = 850 × (-2260)

q₂ = -1 921 000 J

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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

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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.