# A piece of gold initially at a temperature of 25.1 °C absorbs 675 J of heat, raising its temperature to 57.4 °C. Assuming the specific heat of gold is 0.126 J/(g°C), what is the mass of the sample?

Jun 9, 2017

$\text{166 g}$

#### Explanation:

The key here is the specific heat of gold, which is said to be equal to

${c}_{\text{gold" = "0.126 J g"^(-1)""^@"C}}^{- 1}$

This tells you that in order to increase the temperature of $\text{1 g}$ of gold by ${1}^{\circ} \text{C}$, you need to provide it with $\text{0.126 J}$ of heat.

In your case, the temperature increases from ${25.1}^{\circ} \text{C}$ to ${57.4}^{\circ} \text{C}$, which implies that it changes by

${57.4}^{\circ} \text{C" - 25.1^@"C" = 32.3^@"C}$

Now, you can use the specific heat of gold to calculate how much energy would be needed to increase the temperature of gold by ${32.3}^{\circ} \text{C}$.

32.3 color(red)(cancel(color(black)(""^@"C"))) * "0.126 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "4.0698 J g"^(-1)

This tells you that in order to increase the temperature of $\text{1 g}$ of gold by ${32.3}^{\circ} \text{C}$, you need to provide it with $\text{4.0698 J}$ of heat.

You can thus say that $\text{675 J}$ of heat will increase the temperature of

675 color(red)(cancel(color(black)("J"))) * "1 g"/(4.0698color(red)(cancel(color(black)("J")))) = "165.86 g"

of gold by ${32.3}^{\circ} \text{C}$. Rounded to three sig figs, the answer will be

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of gold = 166 g}}}}$