# What is the specific heat capacity of a metal if it requires 178.1 J to change the temperature to 15.0 g of the metal from 25.00 C to 32.00 C?

Jul 1, 2016

${\text{1.70 J g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

When a problem asks you to find a substance's specific heat, $c$, it's essentially telling you to find how much heat is needed in order to increase the temperature of $\text{1 g}$ of said substance by ${1}^{\circ} \text{C}$.

In your case, you know that you need $\text{178.1 J}$ in order to increase the temperature of $\text{15.0 g}$ of your unknown metal by ${7}^{\circ} \text{C}$, since

$\Delta T = {32.00}^{\circ} \text{C" - 25.00^@"C" = 7.00^@"C}$

So, how much heat would be needed to increase the temperature of $\text{1 g}$ of this metal by ${7.00}^{\circ} \text{C ?}$

1 color(red)(cancel(color(black)("g"))) * "178.1 J"/(15.0color(red)(cancel(color(black)("g")))) = "11.873 J"

Since this much heat is needed to increase the temperature of $\text{1 g}$ by ${7.00}^{\circ} \text{C}$, it follows that you can increase temperature by ${1}^{\circ} \text{C}$ by adding

1color(red)(cancel(color(black)(""^@"C"))) * "11.873 J"/(7.00color(red)(cancel(color(black)(""^@"C")))) = "1.70 J"

Since you need $\text{1.70 J}$ to increase the temperature of $\text{1 g}$ of this metal by ${1}^{\circ} \text{C}$, you can say that the metal's specific heat will be

$c = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{1.70 J g"^(-1)""^@"C}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}} \to$ rounded to three sig figs

ALTERNATIVELY

You can also use the following 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

Rearrange to solve for $c$

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

Plug in your values to find

c = "178.1 J"/("15.0 g" * 7.00^@"C") = color(green)(|bar(ul(color(white)(a/a)color(black)("1.70 J g"^(-1)""^@"C"^(-1))color(white)(a/a)|)))