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Difficulty: Very hardLatent Heat and Changes of State

A copper calorimeter of mass 0.15 kg0.15\text{ kg} contains 0.25 kg0.25\text{ kg} of water at 20.0∘C20.0^\circ\text{C}. Dry steam at 100.0∘C100.0^\circ\text{C} is passed into the water until the final temperature of the mixture reaches 40.0∘C40.0^\circ\text{C}. Assuming no heat is lost to the surroundings, what is the mass of steam condensed, in grams?

(Take specific heat capacity of water =4200 J kg−1 K−1= 4200\text{ J kg}^{-1}\text{ K}^{-1}, specific heat capacity of copper =400 J kg−1 K−1= 400\text{ J kg}^{-1}\text{ K}^{-1}, and specific latent heat of vaporization of water =2.26×106 J kg−1= 2.26 \times 10^6\text{ J kg}^{-1})

Answer: 8.84 g

Answer

8.84 g
By energy conservation, the heat gained by the cold water and copper vessel must equal the total heat released by the condensing steam and the cooling of that condensed water. The heat gained is (0.25 kg × 4200 J/kg·K + 0.15 kg × 400 J/kg·K) × 20.0 K = 22,200 J. The heat lost per kilogram of steam is 2,260,000 J/kg + 4200 J/kg·K × 60.0 K = 2,512,000 J/kg. Dividing 22,200 J by 2,512,000 J/kg gives 0.0088376 kg, which corresponds to 8.84 g.

Step-by-Step Solution

1
Calculate the thermal energy absorbed by the cold water and copper calorimeter
Q_gained = 22,200 J
Both the water and copper calorimeter increase in temperature from 20.0°C to 40.0°C (ΔT = 20.0 K).
2
Express the heat released by mass m_s of steam as it condenses and cools to 40.0°C
Q_lost = m_s * 2,512,000 J
Steam releases latent heat when condensing at 100.0°C (m_s * L_v) and sensible heat when the condensed water cools from 100.0°C to 40.0°C (m_s * c_w * 60.0).
3
Apply the principle of conservation of thermal energy and solve for mass m_s in grams
m_s = 8.84 g
Setting Q_gained equal to Q_lost yields m_s = 22,200 / 2,512,000 = 0.0088376 kg, which equals 8.84 g.

Key Concept

Thermal energy balance involving phase change (latent heat of vaporization) and sensible heat exchange
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Topics

Latent Heat and Changes of State

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