Respuesta :
Let T be the final temperature of water and nails (iron).
Given that the initial temperature of 100 g nails, [tex]T_{1F}=40^{\circ}C[/tex].
Specific heat of the iron, [tex]C_F=0.12 cal/g^{\circ}C[/tex].
So, the heat lost by nails [tex]= mC_F(T_{1F}-T}[/tex]
[tex]=100\times 0.12 \times (40-T)\cdots(i)[/tex]
The initial temperature of 100 g water, [tex]T_{1w}=20^{\circ}C.[/tex]
Specific heat of water, [tex]C_w=1 cal/g^{\circ}C[/tex]
So, the heat gained by water [tex]= mC_w(T-T_{1w}}[/tex]
[tex]=100\times 1 \times (T-20)\cdots(ii)[/tex]
As the exergy is conserved, so
the heat lost by the nails = the heat gained by the water
From equation (i) and (ii), we have
[tex]100\times 0.12 \times (40-T)=100(T-20)[/tex]
[tex]\Rightarrow 0.12(40-T)=T-20[/tex]
[tex]\Rightarrow 0.12\times 40 -120T=T-20[/tex]
[tex]\Rightarrow 0.12T+T=4.8+20[/tex]
[tex]\Rightarrow 1.12T=24.8[/tex]
[tex]\Rightarrow T=24.8/1.12= 22.1^{\circ}C[/tex]
Hence, the final temperature of both, iron nails as well as water, become [tex]22.1^{\circ}C[/tex].
