91Ó°ÊÓ

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:

$m\mathrm{Δ}{C}_p\mathrm{Δ}T$

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

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

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

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

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

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

The initial and final temperatures in Kelvin:

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

The substituting values are:

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

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

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..