# The specific heat of aluminum is 0.214 cal/g°c. What is the energy, in calories, necessary to raise the temperature of a 55.5 g piece of aluminum from 23.0 to 48.6°C?

Jul 19, 2017

$\text{304 cal}$

#### Explanation:

The specific heat of aluminium tells you the amount of energy needed to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$.

${c}_{\text{Al" = "0.214 cal g"^(-1)""^@"C}}^{- 1}$

You can thus say that in order to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$, you need to supply it with $\text{0.214 cal}$ of heat.

Now, you know that your sample has a mass of $\text{55.5 g}$. Use the specific heat of aluminium to calculate how much heat would be needed to increase the temperature of this sample

55.5 color(red)(cancel(color(black)("g"))) * "0.214 cal"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "11.877 cal"""^@"C"^(-1)

So, you now know that in order to increase the temperature of $\text{5.5 g}$ of aluminium by ${1}^{\circ} \text{C}$, you need to supply it with $\text{11.877 cal}$ of heat.

But since you know that the temperature change is equal to

${48.6}^{\circ} \text{C" - 23.0^@"C" = 25.6^@"C}$

you can say that you will need

$25.6 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))) * overbrace("11.877 cal"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 55.5 g of Al")) = color(darkgreen)(ul(color(black)("304 cal}}}}$

The answer is rounded to three sig figs.