# Ten grams of iron cooled from 60.4 °C to 35.0°C, and released 114 J of heat in the process. What is the specific heat of iron?

Jun 25, 2016

${\text{0.45 J g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} q = m \cdot c \cdot \Delta T \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, where

$q$ - the amount of heat released
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature

Now, because your sample of iron is being cooled, the heat it released will actually carry a negative sign.

$q = - \text{114 J}$

That happens because the final temperature of the sample is lower than the initial temperature, which makes the change in temperature negative.

You will have

$\Delta T = {35.0}^{\circ} \text{C" - 60.4^@"C" = -25.4^@"C}$

All you have to do now is rearrange the equation to solve for $c$, the specific heat of iron

$q = m \cdot c \cdot \Delta T \implies c = \frac{q}{m \cdot \Delta T}$

Plug in your values to find

c = (-"114 J")/("10 g" * (-25.4^@"C")) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.45 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))

I'll leave the answer rounded to two sig figs.

Iron's specific heat is listed as being equal to ${\text{0.44 J g"^(-1)""^@"C}}^{- 1}$, so this is an excellent result.

http://www2.ucdsb.on.ca/tiss/stretton/database/Specific_Heat_Capacity_Table.html