How many grams of iron at 200.0 c must be placed in 200.0 g of water at 18.0 c so that the temp of both will be 30.0 c ?
Okay so the SH of iron = .106
This is what I have set up.. Not sure if I'm going at this the right way..
(.106) (m) (170.0) = (1.0) (200.0) (12.0)
I don't know how I would find the grams needed.. This is what I have set up
2 answers
I got 750 g but I believe my set up is wrong
I assume that 0.106 value has units of cal/g. All of the numbers you put in are right although I would arrange them differently. I don't believe 750 g is right. See below for my explanation.
heat lost by iron + heat gained by H2O = 0
[mass Fe x s.h. Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
[mass Fe x 0.106 x (30-200)] + [200 x 1.0 x (30-18) = 0 which gives you
(mass Fe x 0.106 x -170) + (200 x 1.0 x 12) = 0, then solve for mass Fe.
heat lost by iron + heat gained by H2O = 0
[mass Fe x s.h. Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0
[mass Fe x 0.106 x (30-200)] + [200 x 1.0 x (30-18) = 0 which gives you
(mass Fe x 0.106 x -170) + (200 x 1.0 x 12) = 0, then solve for mass Fe.
1 answer