Question

Question #4719e

Answer 1

`"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")))`