chemistry

A 32.2 g iron rod, initially at 22.3 C, is submerged into an unknown mass of water at 63.9 C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 58.6 C.

What is the mass of the water?

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  1. heat gained by rod + heat lost by H2O = 0

    [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

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  2. when I work this out I get 5.7 g but it is saying that my answer is incorrect and I am not sure what I am doing wrong..

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  3. Post your work and i'll find the error. The answer is about 25 g water.

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  4. Going off line but if you post your work I'll look at it first thing tomorrow.

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  5. heat gained by rod + heat lost by H2O = 0

    [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

    32.2*0.450*(58.4-22.8)+m*4.179*(58.4-63.7)=0

    515.844=m*22.148

    m=23.3 grams

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  6. heat gained by rod + heat lost by H2O = 0

    [mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

    32.2*0.450*(58.6-22.3)+m*4.179*(58.6-63.9)=0

    525.987=m*22.148

    m=23.748 grams

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  7. you are supposed to round to 2 sig figs.. answer is 24g

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