Question

Question #2afc8

Answer 1

Here's what I got.

Explanation:

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

`c_"Al" = "0.91 J g"^(-1)""^@"C"^(-1)`

`c_"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^@"C " -> " "19^@"C"`

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

`17^@"C " -> " " 19^@"C"`

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

In your case, the heat released when the temperature of `"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^@"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 `"7,280 J"` as it cools from `99^@"C"` to `19^@"C"`. This implies that the unknown mass of water will absorb `"7,280 J"` of heat as it heats by

`19^@"C" - 17^@"C" = 2^@"C"`

Since you know that you need `"4.18 J"` of heat to increase the temperature of `"1 g"` of water by `1^@"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^@"C"`. This implies that `"7,280 J"` will be enough to increase the temperature of

`"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^@"C"`. Therefore, you can say that the mass of water present in the container is equal to

`color(darkgreen)(ul(color(black)("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.