# How do I find the specific heat in this question?

## When a 57-gram piece of aluminum at 100oC is placed in water, it loses 735 calories of heat while cooling to 30oC. Calculate the specific heat of the aluminum

Mar 12, 2017

${c}_{\text{Al" = "0.184 cal g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

The specific heat of aluminium tells you the amount of heat needed or released when the temperature of $\text{1 g}$ of aluminium changes by ${1}^{\circ} \text{C}$.

In your case, the temperature decreases by

${100}^{\circ} \text{C" - 30^@"C" = 70^@"C}$

so the value of the specific heat will tell you the amount of heat given off when $\text{1 g}$ of aluminium cools by ${1}^{\circ} \text{C}$.

So, the first thing to do here is to figure out the heat given off when $\text{57 g}$ of aluminium cool by ${1}^{\circ} \text{C}$. In order to do that, use the fact that it $735$ calories were given off when the sample cooled by ${70}^{\circ} \text{C}$

${\text{735 cal"/(70^@"C") = "10.5 cal" ""^@"C}}^{- 1}$

This means that when the temperature of $\text{57 g}$ of aluminium decreases by ${1}^{\circ} \text{C}$, $10.5$ calories of heat are being given off.

Now all you have to do is figure out the amount of heat given off when the temperature of $\text{1 g}$ of aluminium decreases by ${1}^{\circ} \text{C}$

("10.5 cal" ""^@"C"^(-1))/"57 g" = "0.184 cal g"^(-1)""^@"C"^(-1)

And there you have it -- aluminium has a specific heat of

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{c}_{\text{Al" = "0.184 cal g"^(-1)""^@"C}}^{- 1}}}}$

I'll leave the answer rounded to three sig figs, but keep in mind that you only have one significant figure for the initial temperature of the metal.

You can double-check your result by using the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat lost or gained by the substance
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

Now, notice that the change in temperature is actually equal to

$\Delta T = {30}^{\circ} \text{C" - 100^@"C" = - 70^@"C}$

This is consistent with the fact that heat is being lost by the metal. However, you don't really need to use the minus sign when solving for the specific heat of a material.

That is the case because specific heat is meant to symbolize both heat absorbed when the temperature of $\text{1 g}$ increases by ${1}^{\circ} \text{C}$ and heat given off when the temperature of $\text{1 g}$ decreases by ${1}^{\circ} \text{C}$.

In this regard, specific heat is not meant to indicate if heat is being lost or gained. Therefore, when calculating the specific heat of a material, you can go for

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

This will ensure that the specific heat comes out positive regardless if the material loses or gains heat.

So, plug in your values to find

c = "735 cal"/("57 g" * |-70^@"C"|) = color(darkgreen)(ul(color(black)("0.184 cal g"^(-1)""^@"C"^(-1))))