1.5 kg of ice melts  when a jet of steam at 100 degree C is passed in an ice block .Calculate the mass of steam required to achieve it

Asked by lovemaan5500 | 29th Dec, 2017, 06:39: AM

Expert Answer:

Let MS be the required mass of steam in kg.
 
Heat loss by the steam from 100°C to 0C is,  QL = MS×Lv+MS×Cp×(100-0) 
 
where Lv is latent heat of vaporaisation of steam, Cp is specific heat of water
 
Heat required to Melt 1.5 kg of ice, QG = 1.5×Lf
where Lf is latent heat of vaporaisation of steam
 
QL = QG
 
MS×Lv+MS×Cp×(100-0)  = 1.5×Lf
 
 MS = 1.5 × [ Lf / ( Lv + 100 × Cp ) ]
 
Lf = 334 kJ/kg, Lv = 2260 kJ/kg and Cp = 4.2 kJ/(kgK)
 
MS = ( 1.5×334 ) /2680 = 187 gm

Answered by  | 29th Dec, 2017, 12:37: PM