Question #69369

1 Answer
Jan 10, 2017

Answer:

Here's why that is the case.

Explanation:

The trick here is to realize that both beakers contain the same amount of hydrochloric acid, but different amounts of water.

The reaction between magnesium and hydrochloric acid forms magnesium chloride and hydrogen gas

#"Mg"_ ((s)) + 2"HCl"_ ((aq)) -> "MgCl"_ (2(aq)) + "H"_ (2(g)) uarr#

The important thing to mention here is that the reaction is exothermic, which means that it gives off heat as it progresses.

Notice that beaker #"A"# contains

#100 color(red)(cancel(color(black)("cm"^3))) * (1 color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 mole HCl"/(1color(red)(cancel(color(black)("dm"^3)))) = "0.10 moles HCl"#

Similarly, beaker #"B"# contains

#200 color(red)(cancel(color(black)("cm"^3))) * (1 color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "0.5 moles HCl"/(1color(red)(cancel(color(black)("dm"^3)))) = "0.10 moles HCl"#

You know that the two pieces of magnesium are identical, which means that both reactions will give off the same amount of heat.

Now, this heat is being off by the reaction to its surroundings, i.e. to the solution.

The temperature of the solution in beaker #"A"# will indeed be higher than the temperature of the solution in beaker #"B"# because the former contains less water than the latter.

In this regard, it requires less heat to get to a higher temperature than beaker #"B"#. The amount of heat is the same, so the beaker that holds less water will get to a higher temperature than the beaker that holds more water.

In other words, a smaller mass of water requires less heat to get to the same temperature as a bigger mass of water.

If you take #m# to be the mass of water in beaker #"A"# and #2m# to be the mass of water in beaker #"B"#, you can say that

#{(q = m * c * DeltaT_"A"), (q = 2m * c * DeltaT_"B") :}#

Here

  • #q# is the heat absorbed by the water, the same for beaker #"A"# and for beaker #"B"#
  • #c# is the specific heat of water
  • #DeltaT_"A"# is the change in temperature for beaker #"A"#
  • #DeltaT_"B"# is the change in temperature for beaker #"B"#

Notice that you have

#color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * DeltaT_"A" = 2color(red)(cancel(color(black)(m))) * color(red)(cancel(color(black)(c))) * DeltaT_"B"#

which gets you

#DeltaT_"A" = 2 * DeltaT_"B"#

This tells you that the increase in temperature is twice as high for the solution in beaker #"A"# than for the solution in beaker #"B"#.