Question

Question #cf198

Answer 1

`"0.521 J g"^(-1)""^@"C"^(-1)`

Explanation:

The idea here is that the heat given off by the metal is equal to the heat absorbed by the water.

`color(blue)(ul(color(black)(-q_"metal" = q_"water")))`

The minus sign is used here because heat given off carries a negative sign.

The first thing to do here is to convert the volume of water to mass by using water's density. At `20.0^@"C"`, water has a density of

`rho ~~ "0.99821 g mL"^(-1)`

http://antoine.frostburg.edu/chem/senese/javascript/water-density.html

This means that your sample has a mass of

`100.0 color(red)(cancel(color(black)("mL"))) * "0.99821 g"/(1color(red)(cancel(color(black)("mL")))) = "99.821 g"`

Now, water has a specific heat of

`c_"water" = "4.18 J g"^(-1)""^@"C"^(-1)`

To figure out how much heat was absorbed by the water, use the equation

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

Here

• `q` is the heat absorbed / given off
• `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

In your case, the sample goes from `20.0^@"C"` to `24.5^@"C"`, which means that

`DeltaT_ "water" = 24.5^@"C" - 20.0^@"C" = 4.5^@"C"`

Plug your values into the above equation to find the amount of heat absorbed by the water

`q_"water" = 99.821 color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 4.5 color(red)(cancel(color(black)(""^@"C")))`

`q_"water" = "1877.6 J"`

This is exactly how much heat was given off by the metal when it cooled from `150.0^@"C"` to `24.5^@"C"`. The change in temperature for the metal will be

`DeltaT_"metal" = 24.5^@"C" - 150.0^@"C" = -125.5^@"C"`

Keep in mind that

`color(blue)(ul(color(black)(-q_"metal" = q_"water")))`

which means that you have

`q_"metal" = - "1877.6 J"`

Rearrange the equation to solve for `c_"metal"`

`q_"metal" = m_"metal" * c_"metal" * DeltaT_"metal"`

`c_"metal" = q_"metal"/(m_"metal" * DeltaT_"metal")`

Plug in your values to find

`c_"metal" = (color(red)(cancel(color(black)(-)))"1877.6 J")/("28.7 g" * (color(red)(cancel(color(black)(-))) 125.5^@"C")`

`c_"metal" = color(darkgreen)(ul(color(black)("0.521 J g"^(-1)""^@"C"^(-1)))`

The answer is rounded to three sig figs.