# 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?

##### 1 Answer

#### Answer:

#### Explanation:

Your tool of choice here will be the equation

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*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 = -"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

#DeltaT = 35.0^@"C" - 60.4^@"C" = -25.4^@"C"#

All you have to do now is rearrange the equation to solve for

#q = m * c * DeltaT implies c = q/(m * DeltaT)#

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

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