# Question 2afc8

Jun 15, 2017

Here's what I got.

#### Explanation:

Start by looking up the specific heats of aluminium and of water

${c}_{\text{Al" = "0.91 J g"^(-1)""^@"C}}^{- 1}$

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

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html

The idea here is that the heat lost by the block of aluminium as its temperature decreases

${99}^{\circ} \text{C " -> " "19^@"C}$

will be equal to the heat absorbed by the water as its temperature increases

${17}^{\circ} \text{C " -> " " 19^@"C}$

Now, aluminium has a specific heat of ${\text{0.91 J g"^(-1)""^@"C}}^{- 1}$, which tells you that in order to increase the temperature of $\text{1 g}$ of aluminium by ${1}^{\circ} \text{C}$, you need to provide it with $\text{0.91 J}$ of heat.

In your case, the heat released when the temperature of $\text{100 g}$ of aluminium decreases is equal to

100 color(red)(cancel(color(black)("g"))) * "0.91 J"/(1color(red)(cancel(color(black)("g"))) * 1^@"C") = "91 J"^@"C"^(-1)

In your case, the temperature must decrease by

${99}^{\circ} \text{C" - 19^@"C" = 80^@"C}$

which means that the block will release

80color(red)(cancel(color(black)(""^@"C"))) * overbrace("91 J"/(1color(red)(cancel(color(black)(""^@"C")))))^(color(blue)("for 100 g of Al")) = "7,280 J"

So, you know that your block of aluminium will give off $\text{7,280 J}$ as it cools from ${99}^{\circ} \text{C}$ to ${19}^{\circ} \text{C}$. This implies that the unknown mass of water will absorb $\text{7,280 J}$ of heat as it heats by

${19}^{\circ} \text{C" - 17^@"C" = 2^@"C}$

Since you know that you need $\text{4.18 J}$ of heat to increase the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$, you can say that you will need

2 color(red)(cancel(color(black)(""^@"C"))) * "4.18 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "8.36 J g"^(-1)#

to increase the temperature of water by ${2}^{\circ} \text{C}$. This implies that $\text{7,280 J}$ will be enough to increase the temperature of

$\text{7,280" color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(8.36color(red)(cancel(color(black)("J")))))^(color(blue)("for a 2-"""^@"C increase")) = "870.8 g}$

of water by ${2}^{\circ} \text{C}$. Therefore, you can say that the mass of water present in the container is equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{mass of water = 870 g}}}}$

I'll leave the answer rounded to two sig figs, but keep in mind that you only have one significant figure for the mass of the block.