Answer:
The temperature will be "338 K".
Explanation:
The given values are:
[tex]c_{Ag}=233 \ J/kg-K[/tex]
Temperature, [tex]\Delta \ T=263 \ K[/tex]
As we know,
⇒  [tex]Heat \ gain \ by \ ice = Heat \ last \ by \ silver[/tex]
[tex]m_{ice}c_{ice} \Delta \ T+m_{ice}L_{ice}+m_{ice}c_{w} \Delta \ T=m_{Ag}c_{Ag} \Delta \ T[/tex]
On putting the values, we get
[tex]Q_{ice}=4.7\times 2030(273-263)+4.7(330\times 10^3)+4.7\times 4186(TK-273K)[/tex]
⇒   [tex]=-3724646.6 \ J+19674.2 \ T[/tex]
[tex]Q_{Ag}=16.7\times 233(1087-T)[/tex]
⇒   [tex]=4229625.7-3891.1 \ T[/tex]
Now,
System's final temperature will be:
                 [tex]Q_{ice}=Q_{Ag}[/tex]
[tex]-3724646.6 \ J+19674.2 \ T=4229625.7-3891.1 \ T[/tex]
                   [tex]T=338 \ K[/tex]
