Answer to Question #306779 in Chemistry for Aleeha

Question

calculate the amount of energy required or realeased to cool down 12.5g of steam from 125.0 degree celsius to ice at -20 degree celsius

Answer 1

Specific heat capacity of steam = С_steam = 2.01 J g⁻¹ °C⁻¹

Specific heat capacity of water = C_water = 4.186 J g⁻¹ °C⁻¹

Specific heat capacity of ice = C_ice = 2.10 J g⁻¹ °C⁻¹

Latent heat of condensation of water = ∆H_con = −2256 J/g

Latent heat of crystallization of water = ∆H_cry = −333.7 J/g

Solution:

This problem requires several steps:

Step 1: cool 12.5 g of steam from 125°C to 100°C

q₁ = m × С_steam × ∆T = (12.5 g) × (2.01 J g⁻¹ °C⁻¹) × (100°C − 125°C) = −628.125 J

q₁ = −628.125 J

Step 2: convert 12.5 g of steam at 100°C to 12.5 g of water at 100°C

q₂ = m × ∆H_con = (12.5 g) × (−2256 J/g) = −28200 J

q₂ = −28200 J

Step 3: cool 12.5 g of water from 100°C to 0°C

q₃ = m × C_water × ∆T = (12.5 g) × (4.186 J g⁻¹ °C⁻¹) × (0°C − 100°C) = −5232.5 J

q₃ = −5232.5 J

Step 4: convert 12.5 g of water at 0°C to 12.5 g of ice at 0°C

q₄ = m × ∆H_cry = (12.5 g) × (−333.7 J/g) = −4171.25 J

q₄ = −4171.25 J

Step 5: cool 12.5 g of ice from 0°C to −20°C

q₅ = m × C_ice × ∆T = (12.5 g) × (2.10 J g⁻¹ °C⁻¹) × (−20°C − 0°C) = −525 J

q₅ = −525 J

Add up all the heat required for all these changes is

q = q₁ + q₂ + q₃ + q₄ + q₅

q = (−628.125 J) + (−28200 J) + (−5232.5 J) + (−4171.25 J) + (−525 J) = −38756.875 J = −38.76 kJ

q = −38.76 kJ

Thus, 38.76 kJ of energy realeased to cool down 12.5g of steam from 125°C to ice at −20°C

Answer: 38.76 kJ of energy realeased