Answer to Question #87211 in Molecular Physics | Thermodynamics for Olivia Bamson

Question

Steam, at 200° C is passed into a container of negligible heat capacity containing 40 g of ice and 200 g of water at 0° C until the ice is completely melted. Determine the total mass of water in the container

Answer 1

Heat required by ice and water to go up to 100°C = m₁L₁ + m₁c₁ Δ T₁ + m₂c₂ Δ T₂

here m₁ = 40 g

L₁ = 80 cal/g ……. ice

C₁ = 1

ΔT₁ = 200 – 0 = 200 °C

m₂ = 200 g

C₂ = 1

ΔT₂ = 100 – 0 = 100 °C

Heat required = 40 × 80 + 200 × 1 × 100 + 200 × 1 × 100 = 43200

Q = 43200 cal

Let m_s is mass of steam

Q = m_sL₂

for water steam L₂ = 540 cal/g

m_s = (Q / L₂)

m_s = [(43200) / (540)]

m_s = 80 g – This is steam converted to water

Total water = 40 + 200 +80 = 320 g