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?

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.

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

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