A 750 g aluminum pan is removed from the stove and plunged

into a sink filled with 10.0 L of water at 20.0C. The water temperature

quickly rises to 24.0C. What was the initial temperature

of the pan in C and in F?

The mass of the pan is $750g,$$10.0L$of water at $20°\mathrm{C}$, and its final temperature is $24.0^{0}c$

formula used:

The equation to calculate the energy of each element is:

Make a solution for$T_{i,\text{aluminum}}:$

$$\begin{gathered} m_{\text{pan}}\Delta C_{p,\text{aluminum}}\Delta T_{i,\text{aluminum}}+m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_{i,\text{water}} \\ =\left(m_{\text{pan}}\Delta C_{p,\text{aluminum}}+m_{\text{water}}\Delta C_{p,\text{water}}\right)\Delta T_f \\ {T}_{i,\text{aluminum}}=\frac{\left(m_{\text{pan}}\Delta C_{\text{p,atuminimen}}+m_{\text{water}}\Delta C_{p,\text{water}}\right)\Delta T_f-m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_{i,\text{water}}}{m_{\text{pan}}\Delta C_{p,\text{Ahiminium}}} \end{gathered}$$

Now, we need to know how much the mass $10\mathrm{L}$of water is:

$10\left[L\right]\Delta \frac{1000\left[g\right]}{1\left[L\right]}=10,000\left[g\right]$

Looking at a table of specific heat capacity, we can find the next information:

$$\begin{gathered} C_{p,a\text{aminum}}=0.897\left[\frac{J}{g\Delta K}\right] \\ {C}_{p,\text{water}}=4.181\left[\frac{J}{g\Delta K}\right] \end{gathered}$$

The initial and final temperatures in Kelvin:

$$\begin{gathered} T_{i,\text{water}}=20\left[{}^{\circ }\mathrm{C}\right]+273.15=293.15\left[K\right] \\ {T}_f=24\left[{}^{\circ }\mathrm{C}\right]+273.15=297.15\left[K\right] \end{gathered}$$

Substituting values:

$T_{i,\text{aluminum}}=\frac{\left(750\left[g\mid \Delta 0.897\left[\frac{J}{g\Delta K}\right]+10,000\left[g\mid \Delta 4.181\left[\frac{J}{g\Delta K}\right]\right)\Delta 297.15\left[K\right]-10,000\left[g\mid \Delta 4.181\left[\frac{J}{g\Delta K}\right]\Delta 293.15\left[K\right]\right.\right.\right.}{750\left|g\right|\Delta 0.897\left[\frac{J}{g\Delta K}\right]}$

$T_{i,\text{aluminum}}=545.74\left[K\right]$

In degrees Celsius:

$T_{i,\text{aluminum}}=545.74\left[K\right]-273.15=272.60\left[{}^{\circ }\mathrm{C}\right]$

In Fahrenheit:

$T_{i,\text{aluminum}}=272.60\left[{}^{\circ }\mathrm{C}\right]\Delta 1.80\left[{}^{\circ }\mathrm{F}\right]+32\left[{}^{\circ }\mathrm{F}\right]=522.67\left[{}^{\circ }\mathrm{F}\right]$

The pan transfers energy to the water until both reach equilibrium, at which point their temperatures are the same. We considered an isolated system in this situation.