Steam at 100 celsius is continuously passed through a flask containing 100 g of ice at -10 degree Celsius to produce boiling water in the flask. heat loss in a flask is prevented, the mass of the boiling water produced in the flask is (use c,ice=0.5 cal g^-1 k^-1 ; c water=1 cal g^-1 k^-1; Lf = 80 cal g^-1 and Lv= 540 cal g^-1 )
A .100g
B. 134g
C.166g
D.66g
Answers
Steam at 100°C is continuously passed through a flask containing 100 g of ice at -10°C to produce boiling water in the flask. heat loss in a flask is prevented.
We have to find the mass of the boiling produced in the flask is ...
“we know that heat flows from higher temperature to lower temperature.”
so, here heat flows from steam to ice and at a certain time both exists in an equilibrium state.
Let the mass of steam is passed through the flask is 'm' g.
∴ heat lost by steam to change its phase into water , H = mLv
= m g × 540 cal/g
= 540m cal
heat gained by ice to increase its temperature from -10°C to 0°C , h₁ = m's'∆T'
= 100g × 0.5 cal/g/°C × (0 - (-10))°C
= 500 cal
heat gained by ice to change its phase into water, h₂ = m'Lf
= 100g × 80 cal/g
= 8000 cal
heat gained by water to increase its temperature from 0°C to 100°C , h₃ = m's∆T
= 100g × 1 cal/g/°C × 100°C
= 10000 cal
∵ heat lost by steam = heat gained by ice
⇒H = h₁ + h₂ + h₃
⇒540m = 500 + 8000 + 10000
⇒m = 18500/540 ≈ 34 g