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Find the mass of 0oC ice that 10 gof 100oC steam will completely melt. Latent heat of fusion of ice is 80cal/g oCand heat of vaporization of water 540 cal/g oC ?

Sagot :

Cricetus

Answer:

The right approach is "80 g".

Explanation:

Given:

[tex]L_f=80 \ Cal/g[/tex]

[tex]L_v=540 \ Cal/g[/tex]

[tex]S=1 \ Cal/g[/tex]

Now,

The amount of heat cooling will be:

= [tex]mL_v+mS \Delta T[/tex]

= [tex](10\times 540)+10\times 1\times (100-0)[/tex]

= [tex]5400+1000[/tex]

= [tex]6400 \ Cal[/tex]

then,

⇒ [tex]m_{ice} L_f=6400[/tex]

[tex]m_{ice}\times 80=6400[/tex]

        [tex]m_{ice}=\frac{6400}{80}[/tex]

                [tex]=80 \ g[/tex]            

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