Question

A thermally insulated pot has 150 g ice at temperature 0°C. How much steam of 100°C has to be mixed to it, so that water of temperature 50°C will be obtained? (Given: latent heat of melting of ice = 80 cal/g, latent heat of vaporization of water = 540 cal/g, specific heat of water = 1 cal/g°C)

Answer 1

Data: m₁ = 150 g,

Δ T₁ = 50°C − 0°C = 50°C, c_w = 1 cal/g·°C,

L₁ = 80 cal/g, L₂ = 540 cal/g,

Δ T₂ = 100°C − 50°C = 50°C, m₂ = ?

Q₁ (heat absorbed by ice) = m₁L₁

= 150 g × 80 cal/g = 12000 cal

Q₂ (heat absorbed by water formed on melting of ice) = m₁ c_w ΔT₁

= 150 g × 1 cal/g·°C × 50°C = 7500 cal

Q₃ (heat given out by steam) = m₂L₂

= m₂ × 540 cal/g

Q₄ (heat given out by water formed on condensation of steam)

= m₂ C_w ΔT₂ = m₂ × 1 cal/g·°C × 50°C

According to the principle of heat exchange,

Q₁ + Q₂ = Q₃ + Q₄

∴ 12000 cal + 7500 cal = m₂ × 540 cal/g + m₂ × 50 cal/g

∴ 19500 cal = m₂ (540 + 50) cal/g

∴ `m_2 = 19500/590`g

= 33.05 g

33.05 g of steam is to be mixed.