Question #2afc8

1 Answer
Jun 15, 2017

Answer:

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.