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
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.
To find the grams of iron needed to raise the temperature of both the iron and water to 30.0°C, you can use the equation for heat transfer:
q = mcΔT
Where:
q = heat absorbed or released (in this case, we want the heat absorbed)
m = mass of the substance (iron in this case)
c = specific heat capacity of the substance (given as 0.106 J/g°C for iron)
ΔT = change in temperature (final temperature - initial temperature)
Let's break down the problem step by step:
Step 1: Calculate the heat absorbed by the water.
Using the equation above, we have:
q_water = m_water * c_water * ΔT_water
Given:
m_water = 200.0 g (mass of water)
c_water = 1.0 J/g°C (specific heat capacity of water)
ΔT_water = (30.0°C - 18.0°C) = 12.0°C
Plugging in the values:
q_water = (200.0 g) * (1.0 J/g°C) * (12.0°C)
q_water = 2400.0 J
Step 2: Calculate the heat absorbed by the iron.
Using the same equation, we have:
q_iron = m_iron * c_iron * ΔT_iron
Given:
c_iron = 0.106 J/g°C (specific heat capacity of iron)
ΔT_iron = (30.0°C - 200.0°C) = -170.0°C (Note: ΔT_iron is negative because the iron is cooling down)
Plugging in the values:
q_iron = (m_iron) * (0.106 J/g°C) * (-170.0°C)
Step 3: Equate the two heat values to find the mass of iron.
We want the heat absorbed by the iron to equal the heat absorbed by the water:
q_iron = q_water
(m_iron) * (0.106 J/g°C) * (-170.0°C) = 2400.0 J
Now, you can solve for the mass of iron (m_iron):
(m_iron) = 2400.0 J / [(0.106 J/g°C) * (-170.0°C)]
Calculate the right side of the equation:
[(0.106 J/g°C) * (-170.0°C)] ≈ -28.82 J/g
Finally:
(m_iron) ≈ 2400.0 J / (-28.82 J/g)
The units of J will cancel out, leaving the mass of iron in grams. Perform the calculation to find the approximate mass of iron needed.