# Question #2e6dd

##### 1 Answer

#### 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 **decreases**, you can say that the metal is **losing heat**.

On the other hand, the water goes from an initial temperature of **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.

#color(blue)(-q_"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

#color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "# , where

*change in temperature*, defined as the difference between the **final temperature** and the **initial temperature**

Water's specific heat can be found here

#c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)#

https://en.wikipedia.org/wiki/Heat_capacity#Mass_heat_capacity_of_building_materials

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

#DeltaT_"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_"water" = "219.7 J"#

This means that the aluminium **must have lost** the exact same amount of heat. Use equation

#q_"metal" = -"219.7 J"#

You know that you have

#q_"metal" = m_"metal" * c_"metal" * DeltaT_"metal"#

The change in temperature for the metal will be

#DeltaT_"metal" = 26.1^@"C" - 100.0^@"C" = -73.9^@"C"#

Rearrange the equation to solve for

#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**.