Question

A 50-g ice cube at 0ºC is submerged into a container of liquid nitrogen. How many kg of nitrogen evaporates if it is at its boiling point of 77K and has a latent heat of vaporization of 200 kJ/kg? The specific heat of ice is 2100 J/(kg ºC).

Answer 1

0.1029 Kg

Step-by-step explanation:

`Given\ data`

`mass\of \ice\ cube =50gm`

`Latent\ heat \of\ vapourization\ of nitrogen(L) =200kJ/kg`

`specific\ heat\ of\ ice=2100 J/(kgc)`

`boiling\ point\ of\ nitrogen=77K\approx-169^(\circ)c`

`now\ Equating\ heat \ absorb\ by \ ice\ from\ liquid\ nitrogen`

`{m_(ice)}c[{\Delta T}]={m_(nitrogen)}L`

`{m_(ice)}2100[0-(-196)}=m_(nitrogen){200* 1000}`

`m_(nitrogen)=0.1029Kg`