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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).

Respuesta :

Answer:

0.1029 Kg

Explanation:

[tex]Given\ data[/tex]

[tex]mass\of \ice\ cube =50gm[/tex]

[tex]Latent\ heat \of\ vapourization\ of nitrogen(L) =200kJ/kg[/tex]

[tex]specific\ heat\ of\ ice=2100 J/(kgc)[/tex]

[tex]boiling\ point\ of\ nitrogen=77K\approx-169^{\circ}c[/tex]

[tex]now\ Equating\ heat \ absorb\ by \ ice\ from\ liquid\ nitrogen [/tex]

[tex]{m_{ice}}c[{\Delta T}]={m_{nitrogen}}L[/tex]

[tex]{m_{ice}}2100[0-(-196)}=m_{nitrogen}{200\times 1000}[/tex]

[tex]m_{nitrogen}=0.1029Kg[/tex]

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