20 g of steam at 100°C is passed into 100 g of ice at 0°C. Find the resultant temperature if latent heat of steam is 540 cal/g…

Q: 20 g of steam at 100°C is passed into 100 g of ice at 0°C. Find the resultant temperature if latent heat of steam is 540 cal/g, latent heat of ice is 80 cal/g and specific heat of water is 1 cal/g°C.

Sol: For steam

Heat lost by the steam in condensation

Q1 = msLs = 20 × 540 = 10800 cal …(1)

For ice

Heat gained by the ice in melting and to rise its temperature from 0°C to 100°C is

Q2 = mice Lice + mice Sw  Δt

= 100 × 80 + 100 × 1 × 100 = 18000 cal …(2)

From eq. (1) and (2); Q2 > Q1

Let θ = resultant temperature of the mixture According to law of method of mixtures Heat lost by steam = Heat gained by ice

msLs + msSwater (100 – θ) = miceLice + miceSwater (θ – 0)

(20 × 540) + 20 × 1 (100 – θ) = (100 × 80) + (100 × 1 × θ)
θ =40°C