Asked by HELP

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?
Answered by DrBob222
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

Answered by HELP
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..
Answered by DrBob222
Post your work and i'll find the error. The answer is about 25 g water.
Answered by DrBob222
Going off line but if you post your work I'll look at it first thing tomorrow.
Answered by Praveen
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
Answered by Praveen
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
Answered by Jon
you are supposed to round to 2 sig figs.. answer is 24g
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