Question

1.You have an unknown amount of water vapor floating in a sealed container at 126℃. After 30 min the container is cooled to -26℃, and a solid sample of water is found at 2.5g. What amount of energy was used for this process to occur?

Answer 1

7.8 kJ. The process used 7.8 kJ.

The container is sealed, so

mass of water vapour = mass of water = mass of ice.

∴ Heat required = heat to cool vapour + heat to condense vapour to water + heat to cool water + heat to freeze water + heat to cool ice

q = q_1 + q_2 + q_3 + q_4 + q_5

Step 1. Calculate q_1

q_1 = mC_1ΔT_1

m = 2.5 g; C_1 = 1.996 J.°C^(-1)g^(-1)

ΔT_1 = T_f – T_i = 100 °C – 126 °C = -26 °C

q_1 = 2.5 g × 1.996 J.°C^(-1)g^(-1) × (-26°C) = -130 J

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

Step 2. Calculate q_2

q_2 = mΔ_cH

Δ_cH = -2257 J·g^(-1)

q_2 = 2.5 g × [-2257 J·g^(-1)] = -5640 J

Step 3. Calculate q_3

q_3 = mC_3ΔT_3

C_3 = 4.182 J.°C^(-1)g^(-1); ΔT_3 = T_f – T_i = 0 °C – 100 °C = -100 °C

q_3 = 2.5 g × 4.182 J.°C^(-1)g^(-1) × (-100 °C) = -1050 J

Step 4. Calculate q_4

q_4 = mΔ_cH

Δ_cH = -334 J·g^(-1)

q_4 = 2.5 g × [-334 J·g^(-1)] = -835 J

Step 5. Calculate q_5

q_5 = mC_5ΔT_5

C_5 = 2.090 J.°C^(-1)g^(-1); ΔT_3 = T_f – T_i = -26 °C –0 °C = -26 °C

q_5 = 2.5 g × 2.090 J.°C^(-1)g^(-1) × (-26 °C) = -136 J

Step 6. Calculate the total heat involved

q = q_1 + q_2 + q_3 + q_4 + q_5

= (-130 – 5640 – 1050 – 835 – 136) J = -7800 J = -7.8 kJ

The process removed 7.8 kJ.