# Question #3c141

##### 1 Answer

#### 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

*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

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,

Use equation

#- 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*,

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