In an industrial process 10 kg of water per hour is to be heated from 20° C to 80° C. To do this steam at 150° C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90° C. How many kg of steam is required per hour? (Specific heat of steam = specific heat of water =1 cal / g ° C, Latent heat of vaporization =540 cal / g )

Question

In an industrial process 10 kg of water per hour is to be heated from 20° C to 80° C. To do this steam at 150° C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90° C. How many kg of steam is required per hour? (Specific heat of steam = specific heat of water =1 cal / g ° C, Latent heat of vaporization =540 cal / g ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Suppose m kg steam is required per hour Heat is released by steam in following three steps (i) When 150° C steam xrightarrow[Q1]100° C steam Q1=m c text steam Δ θ=m × 1(150-100)=50 mcal (ii) When 100° C steam xrightarrow[Q2] 100° C water Q2=m Lv=m × 540=540 mcal (iii) When 100° C water xrightarrow[Q2] 90° C water Q3=m cW Δ θ=m × 1 ×(100-90)=10 mcal Hence total heat given by the steam Q=Q1+Q2+Q3=600 mcal .....(i) Heat taken by 10 kg water Q prime m cW Δ θ=10 × 103 × 1 ×(80-20)=600 × 103 cal Hence Q=Q prime ⇒ 600 m =600 × 103 ⇒ m=103 g =1 kg