# Question cf198

Nov 29, 2016

${\text{0.521 J g"^(-1)""^@"C}}^{- 1}$

#### Explanation:

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{- {q}_{\text{metal" = q_"water}}}}}$

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 ${20.0}^{\circ} \text{C}$, water has a density of

$\rho \approx {\text{0.99821 g mL}}^{- 1}$

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

${c}_{\text{water" = "4.18 J g"^(-1)""^@"C}}^{- 1}$

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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is the heat absorbed / given off
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, the sample goes from ${20.0}^{\circ} \text{C}$ to ${24.5}^{\circ} \text{C}$, which means that

$\Delta {T}_{\text{water" = 24.5^@"C" - 20.0^@"C" = 4.5^@"C}}$

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")))

${q}_{\text{water" = "1877.6 J}}$

This is exactly how much heat was given off by the metal when it cooled from ${150.0}^{\circ} \text{C}$ to ${24.5}^{\circ} \text{C}$. The change in temperature for the metal will be

$\Delta {T}_{\text{metal" = 24.5^@"C" - 150.0^@"C" = -125.5^@"C}}$

Keep in mind that

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{- {q}_{\text{metal" = q_"water}}}}}$

which means that you have

${q}_{\text{metal" = - "1877.6 J}}$

Rearrange the equation to solve for ${c}_{\text{metal}}$

${q}_{\text{metal" = m_"metal" * c_"metal" * DeltaT_"metal}}$

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.