# Solid iron a specific heat of 0.444 J/g°C. If 75 calories of energy were removed from a piece of iron, and the temperature of the iron went from 21.4°C to -25.2°C, what was the mass of the iron?

Jul 21, 2016

When you look at the big picture, what you have is heat flowing out of solid iron and cooling it down.

So, you have to use the heat flow equation:

$\setminus m a t h b f \left(q = m s \Delta T\right)$

where:

• $q$ is the heat flow in $\text{J}$.
• $m$ is the mass of the object in $\text{g}$.
• $s$ is the specific heat capacity of the object in $\text{J/g"cdot""^@ "C}$.
• $\Delta T$ is the change in temperature of the iron in $\text{^@ "C}$.

First, convert the heat from $\text{cal}$ to $\text{J}$. Note that heat flowing out is negatively-signed by convention.

$- 75$ $\cancel{\text{cal" xx "4.184 J"/cancel"cal" = -"313.8 J}}$

Now what you have is:

$\textcolor{b l u e}{{m}_{\text{Fe") = q_"Fe"/(s_"Fe"DeltaT_"Fe}}}$

= q_"Fe"/(s_"Fe"[T_f^"Fe" - T_i^"Fe"])

= (-313.8 cancel("J"))/((0.444 cancel("J")"/g"cdotcancel(""^@ "C"))[(-25.2cancel(""^@ "C")) - (21.4cancel(""^@ "C"))])

$=$ $\textcolor{b l u e}{\text{15.2 g Fe}}$