Question

A 500 g metal sphere is heated to 300C, then dropped into a

beaker containing 300 cm3 of mercury at 20.0C. A short time later

the mercury temperature stabilizes at 99.0C. Identify the metal.

Short answer

The metal is iron:

$C_{p,\text{iron}}=0.450\left[\frac{\mathrm{J}}{\mathrm{g}\Delta \mathrm{K}}\right]$

Step-by-step solution

Given information

We have an unknown metal that has been heated to the temperature of mercury, with a final temperature of$99.0^0$

formula used:

The equation to calculate the energy of each element is:

$m\Delta C_p\Delta T$

Step 2:Calculation

Solving for $C_{p,\text{metal}}:$

$$\begin{gathered} m_{\text{sphere}}\Delta C_{p,\text{metal}}\Delta T_{i,\text{sphere}}-m_{\text{sphere}}\Delta C_{p,\text{metal}}\Delta T_f=m_{\text{mercury}}\Delta C_{p,\text{mercury}}\Delta T_f \\ -m_{\text{mercury}}\Delta C_{p,\text{mercury}}\Delta T_{i,\text{mercury}} \\ {m}_{\text{sphere}}\Delta C_{p,\text{metal}}\left(T_{i,\text{sphere}}-T_f\right)=m_{\text{mercury}}\Delta C_{p,\text{mercury}}\Delta \left(T_f-T_{i,\text{mercury}}\right) \\ {C}_{p,\text{metal}}=\frac{m_{\text{merury}}\Delta C_{p\text{menwry}}\Delta \left(T_f-T_{\text{imereryy}}\right)}{m_{\text{sphere}}\Delta \left(T_{i,\text{sphere}}-T_f\right)} \end{gathered}$$

Now, we need to know how much of the mass $300{\mathrm{cm}}^3$of mercury there is.

$300\left[{\mathrm{cm}}^3\right]\Delta 13.69\left[\frac{\mathrm{g}}{c{m}^3}\right]=4107[\mathrm{g}]$

The following information can be found in a table of specific heat capacity:

$C_{p,\text{mercury}}=0.1395\left[\frac{\mathrm{J}}{\mathrm{g}\Delta \mathrm{K}}\right]$

The initial and final temperatures in Kelvin:

$$\begin{gathered} T_{i,\text{mercury}}=20.0\left[{}^{\circ }\mathrm{C}\right]+273.15=293.15[\mathrm{K}] \\ {T}_{i,\text{sphere}}=300\left[{}^{\circ }\mathrm{C}\right]+273.15=573.15[\mathrm{K}] \\ {T}_f=99.0\left[{}^{\circ }\mathrm{C}\right]+273.15=372.15[\mathrm{K}] \end{gathered}$$

The substituting values are:

$C_{p,metal}=\frac{4107\left|g\right|\Delta 0.1395\left[\frac{J}{g\Delta K}\right]\Delta (372.15[K]-293.15[K\left]\right)}{500\left[g\right]\Delta (573.15[K]-372.15[K\left]\right)}$

$C_{p,\text{metal}}=0.450\left[\frac{\mathrm{J}}{\mathrm{g}\Delta \mathrm{K}}\right]$

We can see that it relates to iron in a table of specific heat.

As we can see, these equations can even aid in the identification of materials..