Question

What will be the final temperature of water that results when 5g of steam at 100°C is bubbled into 500g of water initially at 21°C

Answer 1

Latent heat of vaporization ΔH_vap = 2260 J/g

Specific heat of water c = 4.186 J/g*C

Q₁ = Q₂

Where Q₁ is heat lost by steam

Q_{2 is} heat gained by water

Q₁ = m₁ΔH_vap + cm₁ (T₂ - T₁)

Q₂ = cm₂(T₂ - T₁)

m₁ΔH_vap + cm₁ (T₂ - T₁) = Q₂ = cm₂(T₂ - T₁)

As for the steam T₂<T₁ the value m₁ΔH_vap + cm₁ (T₂ - T₁) will be negative, but as we should add m₁ΔH_vap + cm₁ (T₂ - T₁), we should put "-" before m₁ΔH_vap + cm₁ (T₂ - T₁)

m₁ΔH_vap - cm₁ (T₂ - T₁) = Q₂ = cm₂(T₂ - T₁)

5g * 2260J/g - 4.186 J/g*C * 5 g *(T₂ - 100C) = 4.186 J/g*C * 500 g *(T₂ - 21C)

11300 - 20.93*T₂ + 2093 = 2093*T₂ - 43953

57346 = 2113.93* T₂

T₂ = 27C

Answer: 27 C