# Question 2e6dd

Mar 25, 2016

$\text{3.30 g}$

#### Explanation:

Start by comparing the final temperature of the water + aluminium sample with the initial temperatures of the two substances.

Notice that aluminium goes from an initial temperature of ${100.0}^{\circ} \text{C}$ to a final temperature of ${26.1}^{\circ} \text{C}$. Since the temperature of the metal decreases, you can say that the metal is losing heat.

On the other hand, the water goes from an initial temperature of ${23.7}^{\circ} \text{C}$ to a final temperature of ${26.1}^{\circ} \text{C}$. Since the temperature of the water is Increasing, you can say that it's gaining heat.

The idea here is that the heat lost by the metal will be equal to the heat gained by the water.

$\textcolor{b l u e}{- {q}_{\text{metal" = q_"water")" " " "color(orange)("(*)}}}$

The Minus sign is used here because heat lost will always carry a negative sign.

Now, in order to determine how much heat was lost / gained, you can use the 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 gained / lost
$m$ - 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

Water's specific heat can be found here

${c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}$

https://en.wikipedia.org/wiki/Heat_capacityMass_heat_capacity_of_building_materials

So, you know the mass of water, its specific heat, and its change in temperature

$\Delta {T}_{\text{water" = 26.1^@"C" - 23.7^@"C" = 2.40^@"C}}$

so plug in your values and find how much heat was absorbed by the water

q_"water" = 21.9 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * 2.4color(red)(cancel(color(black)(""^@"C")))

${q}_{\text{water" = "219.7 J}}$

This means that the aluminium must have lost the exact same amount of heat. Use equation $\textcolor{\mathmr{and} a n \ge}{\text{(*)}}$ to get

${q}_{\text{metal" = -"219.7 J}}$

You know that you have

${q}_{\text{metal" = m_"metal" * c_"metal" * DeltaT_"metal}}$

The change in temperature for the metal will be

$\Delta {T}_{\text{metal" = 26.1^@"C" - 100.0^@"C" = -73.9^@"C}}$

Rearrange the equation to solve for ${m}_{\text{metal}}$ and plug in your values to find

m_"metal" = q_"metal"/(c_"metal" * DeltaT_"metal")

m_"metal" = (color(blue)(cancel(color(black)(-)))219.7color(red)(cancel(color(black)("J"))))/(0.900color(red)(cancel(color(black)("J")))/("g" * color(red)(cancel(color(black)(""^@"C")))) * (color(blue)(cancel(color(black)(-)))73.9color(red)(cancel(color(black)(""^@"C"))))) = color(green)(|bar(ul(color(white)(a/a)"3.30 g"color(white)(a/a)|)))

The answer is rounded to three sig figs.