Question

Question #3c141

Answer 1

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

Explanation:

The idea here is that the metal + water system reaches thermal equilibrium because the heat lost by the metal will be equal to the heat gained by the water.

`color(blue)(-q_"metal" = q_"water")" " " "color(orange)("(*)")`

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

So, your tool of choice here will be this equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)))|)" "`, where

`q` - the amount of heat gained / lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

The change in temperature for the two substance will depend on their initial temperature and on the equilibrium temperature

`"For the metal: " DeltaT_"metal" = 27.8^@"C" - 100.0^@"C" = -72.2^@"C"`

`"For water: " DeltaT_"water" = 27.8^@"C" - 23.7^@"C" = 4.1^@"C"`

Notice that the problem provides you with the volume of water, but that you need to know its mass.

To do that, use water's density at `23.7^@"C"`, which is approximately equal to `"0.9974 g mL"^(-1)`.

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

The mass of the sample will thus be

`100.0color(red)(cancel(color(black)("mL"))) * "0.9974 g"/(1color(red)(cancel(color(black)("mL")))) = "99.74 g"`

The specific heat of water is equal to

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

You now have everything that you need in order to find the specific heat of the metal, `c_"metal"`.

Use equation `color(orange)("(*)")` to get

`- m_"metal" * c_"metal" * DeltaT_"metal" = m_"water" * c_"water" * DeltaT_"water"`

`-"59.05 g" * c_"metal" * (-72.2^@"C") = 99.74color(red)(cancel(color(black)("g"))) * "4.18 J" color(red)(cancel(color(black)("g"^(-1))))color(red)(cancel(color(black)(""^@"C"^(-1)))) * 4.1color(red)(cancel(color(black)(""^@"C")))`

Rearrange to find

`c_"metal" = (99.74 * 4.18)/(55.09 * 72.2) "J"/("g" ""^@"C")`

`c_"metal" = "0.10482 J g"^(-1)""^@"C"^(-1)`

Rounded to three sig figs, the number of sig figs you have for the initial temperature of the water and for the equilibrium temperature, the answer will be

`c_"metal" = color(green)(|bar(ul(color(white)(a/a)"0.105 J g"^(-1)""^@"C"^(-1)color(white)(a/a)))|)`

A really good candidate for this unknown metal is iron, `"Fe"`, which has a specific heat of `"0.0108 J g"^(-1)""^@"C"^(-1)`.

http://www.engineeringtoolbox.com/specific-heat-metals-d_152.html