# Question b29dc

Feb 2, 2016

$\Delta T = {38}^{\circ} \text{C}$

#### Explanation:

The key to understanding this type of problems lies with the substance's specific heat.

For a given substance, its specific heat will tell you how much heat is needed in order to increase the mass of $\text{1 g}$ of that substance by ${1}^{\circ} \text{C}$.

In your case, the specific heat of gold is said to be

c = 0.128"J"/("g" ""^@"C")

This tells you that you need $\text{0.128 J}$ of heat for every gram of the sample and for every ${1}^{\circ} \text{C}$ increase in its temperature!

So, for example, your sample of gold is said to have a mass of $\text{20 g}$. What if we wanted to increase the temperature of this sample by just 1^@"C"?

Well, since you need $\text{0.128 J}$ for every gram, and since you only want to increase its temperature by 1^@C", it follows that you'd need

$20 \times \text{0.128 J" = "2.56 J}$

But you ended up using $\text{96 J}$, which means that the difference between this value and the value given to you went into increasing the sample's temperature!

You can say that

$x \times \text{2.56 J" = "96 J}$

Here $x$ represents the increase in temperature produced by adding that extra heat to the sample. This means that you have

$x = \frac{96}{2.56} = 37.5$

The temperature of the coin increased by ${37.5}^{\circ} \text{C}$.

You actually have an equation that you can use for such problems

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat added / removed
$m$ - the mass of the sample
$c$ - the specific heat of the substance
$\Delta T$ - the change in temperature

Plugging your values into this equation will once again give you

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

DeltaT = (96 color(red)(cancel(color(black)("J"))))/(20 color(red)(cancel(color(black)("g"))) * 0.128color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 37.5^@"C"#

I'll leave the answer rounded to two sig figs

$\Delta T = \textcolor{g r e e n}{{38}^{\circ} \text{C}}$