Answer: [tex]38.5^0C[/tex]
Explanation:
To calculate the initial temperature of the water:
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat absorbed = [tex]5.73\times 10^5J[/tex]
[tex]c[/tex] = specific heat of water = [tex]4.18J/g^oC[/tex]
m = mass of water = 2230 g
[tex]T_{final}[/tex] = final temperature of water = [tex]100^0C[/tex]
[tex]T_{initial}[/tex] = initial temperature of metal = ?
Now put all the given values in the above formula, we get:
[tex]5.73\times 10^5J=2230g\times 4.18J/g^oC\times (100-T_i)^0C[/tex]
[tex](100-T_i)=61.5[/tex]
[tex]T_i=100-61.5)=38.5^0C[/tex]
Thus, the initial temperature of the water is [tex]38.5^0C[/tex]
