Question #4719e

1 Answer
Mar 15, 2017

Answer:

#"69.0 g"#

Explanation:

Your tool of choice here will be this equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

  • #q# is the heat lost or gained by the substance
  • #m# is the mass of the sample
  • #c# is the specific heat of the substance
  • #DeltaT# is the change in temperature, defined as the difference between the final temperature and the initial temperature of the sample

As you know, the specific heat of copper is listed as

#c_"copper" = "0.385 J g"^(-1)""^@"C"^(-1)#

http://www2.ucdsb.on.ca/tiss/stretton/database/specific_heat_capacity_table.html

So, the problem wants you to determine the mass of copper that would undergo a #35.5^@"C"# increase in temperature upon the addition of #"943.06 J"#.

The temperature of the sample increases by #35.5^@"C"#, so you can say that

#DeltaT = 35.5^@"C"#

Rearrange the equation to isolate #m# on one side

#q = m * c * DeltaT implies m = q/(c * DeltaT)#

Plug in your values to find

#m = (943.06 color(red)(cancel(color(black)("J"))))/(0.385 color(red)(cancel(color(black)("J"))) "g"^(-1) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 35.5 color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("69.0 g")))#

The answer is rounded to three sig figs.

This means that if you supply #"943.06 J"# of heat to #"69.0 g"# of copper, you will cause its temperature to increase by #35.5^@"C"#.

#color(white)(a/a)#

ALTERNATIVE APPROACH

You can get the same result by using the specific heat of the metal, which tells you that in order to increase the temperature of #"1 g"# of copper by #1^@"C"#, you need to supply it with #"0.385 J"# of energy.

Start by calculating the amount of energy needed to increase the temperature of #"1 g"# of copper by #35.5^@"C"#

#35.5 color(red)(cancel(color(black)(""^@"C"))) * "0.385 J"/("1 g" * 1color(red)(cancel(color(black)(""^@"C")))) = "13.6675 J g"^(-1)#

Since you know that you have #"943.06 J"# available, you can use this value to find the number of grams of copper

#943.06 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(13.6675color(red)(cancel(color(black)("J")))))^(color(blue)("needed for 35.5"^@"C increase in temperature")) = color(darkgreen)(ul(color(black)("69.0 g")))#