A 45.5 g sample of iron is cooled in a 0.750 kg water bath originally at 22.00*C. Both substances reach a final temperature of 23.40*C. Assuming the heat is transferred only between the water and iron, what was the initial temperature of the iron?

Question

A 45.5 g sample of iron is cooled in a 0.750 kg water bath originally at 22.00*C.  Both substances reach a final temperature of 23.40*C.  Assuming the heat is transferred only between the water and iron, what was the initial temperature of the iron?

anonymous · Answer

The Specific heat of Fe(s) is 0.4494 J/gx*C and H20(l) is 4.184  J/gx*C

accessdenied · Answer

These two formulas may help us
   $ \rm q = m \times C_p \times \Delta T = m \times C_p \times \left( T_{final} - T_{initial} \right) $
   $ \rm q_{sys} = - q_{surr} $
The heat lost by the piece of iron is equal to the heat gained by the water.

We have the information to calculate the heat gained by the water using the first formula listed.  Mass (m, grams) times Specific Heat (C_p,  Joules per gram Celsius deg.) times temperature change (Delta T, Celsius degrees).  This gives us the heat, in Joules, gained by the water.

Then, noting that the same heat is lost by the iron, we can use the first equation listed again this time plugging in q, m, specific heat, and final temperature (same as that of water).  We only need Algebra to solve for the initial temperature!