# Question #cf198

##### 1 Answer

#### Answer:

#### 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

#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 thechange in temperature, defined as the difference between thefinal temperatureand theinitial temperature

In your case, the sample goes from

#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

#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

#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**.