Answer : The final equilibrium temperature of the water and iron is, 537.12 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
CIO4-=-1
CI=4O=-1
O has a 2- oxidation change so
CI+4(-2)=-1
CI+(-8)=-1
CI=-1+8=7
So the oxidation number of chlorine is 7 in this case
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