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]
