Question

30 g of copper pellets are removed from a 300C oven and immediately

dropped into 100 mL of water at 20C in an insulated

cup. What will the new water temperature be?

Short answer

The water temperature of $T_f=300.67\left[K\right]=27.52\left[{}^{\circ }\mathrm{C}\right]$$T_f=300.67\left[K\right]=27.52\left[{}^{\circ }\mathrm{C}\right]$

Step-by-step solution

Step 1:Given information

$30\mathrm{g}$of copper pellets at ${300}^{0}c$and $100ML$of water at ${20}^{0}c$

formula used:

The formula for calculating each element's energy is:$m\Delta C_p\Delta T$

Calculation

Solving for $T_f$:

$$\begin{gathered} m_{\text{copper}}\Delta C_{p,\text{copper}}\Delta T_{i,\text{copper}}+m_{\text{water}}\Delta C_{p,\text{water}}\Delta T_{i,\text{water}} \\ =\left(m_{\text{copper}}\Delta C_{p,\text{copper}}+m_{\text{water}}\Delta C_{p,\text{water}}\right)\Delta T_f \\ {T}_f=\frac{m_{\text{copper}}\Delta C_{\text{peopper}}\Delta T_{\text{i, copper}}+m_{\text{wate}}\Delta C_{p,\text{watere}}\Delta T_{\text{i,water}}}{m_{\text{copper}}\Delta C_{\text{pcopper}}+m_{\text{water}}\Delta C_{p,\text{water}}} \end{gathered}$$

Now we need to figure out what kind of mass there is. $100ML$The water is $100[\mathrm{mL}]\Delta \frac{1|\mathrm{g}|}{1[\mathrm{mL}\mid }=100[\mathrm{g}]$

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

$C_{p,\text{copper}}=0.385\left[\frac{\mathrm{J}}{\mathrm{g}\Delta \mathrm{K}}\right]$

$C_{p,\text{water}}=4.1813\left[\frac{\mathrm{J}}{\mathrm{g}\Delta \mathrm{K}}\right]$

The initial temperature in Kelvin:

$$\begin{gathered} T_{i,\text{copper}}=300\left[{}^{\circ }\mathrm{C}\right]+273.15=573.15[\mathrm{K}] \\ {T}_{i,\text{water}}=20\left[{}^{\circ }\mathrm{C}\right]+273.15=293.15[\mathrm{K}] \end{gathered}$$

Substituting values:

$T_f=\frac{30|\mathrm{g}|\Delta 0.385\left[\frac{\mathrm{J}}{8\Delta K}\right]\Delta 573.15\left[K\right]+100|\mathrm{g}|\Delta 4.181\left[\frac{\mathrm{J}}{g\Delta K}\right]\Delta 293.15\left[K\right]}{30|\mathrm{g}|\Delta 0.385\left[\frac{\mathrm{J}}{8\Delta K}\right]+100|\mathrm{g}|\Delta 4.181\left[\frac{\mathrm{J}}{8\Delta K}\right]}$

$T_f=300.67\left[K\right]=27.52\left[{}^{\circ }\mathrm{C}\right]$

The energy that the copper pellets lose is replaced by the energy gained by the water. There is no energy leakage into the environment because it is an isolated system.