Answer:
[tex]a.T_f=100.00\textdegree C\\\\b. 211\ grams\\[/tex]
Explanation:
a.
-The specific heat of iron is 0.45 J /g/K
-Heat released by red hot iron to cool to 100°C
[tex]=130g \times 0.45J/g/K\times 645\textdegree C\\\\=37732.5J[/tex]
#The heat required by water to heat up to 100°C :
[tex]= 85g \times 4.2 \times (100-20)\textdegree C\\\\ = 28560 J[/tex]
#You notice that this heat is less that heat supplied by the iron slug so the equilibrium temperature will be 100 ° C.
#Let m be the grams of water vaporized during the prices:
[tex]H_v=m \times 540x4.2\\ \\ = 2268m J[/tex]
#Heat required to warm the water of 85 g to 100 °C is calculated as:
[tex]H_w=85X4.2 X 80\\ \\= 28560 J\\[/tex]
#For an equilibrium:
Heat Lost= Heat Gained
[tex]37732.5J=28560J+2268m \ J\\\\m=4 \ g[/tex]
Hence, the amount of water vaporized is 4 grams
b. Final mass of iron and remaining water is calculated as;
[tex]M_b=M_b-m\\\\=81g+130g-4g\\\\=211g\\[/tex]
Hence, the final mass of the iron and the remaining water is 211 grams.
