# Heat transfer problem

For this problem, I'm wondering if the thermos should be involved in the heat transfer because right we have an equation in calorimetry: $q_\text{warm} =-(q_\text{cold} + q_\text{cal})$

A 125 g sample of cold water and a 283 g sample of hot water are mixed in an insulated thermos bottle and allowed to equilibrate. If the initial temperature of the cold water is 3.0 °C, and the initial temperature of the hot water is 91.0 °C, what will be the final temperature?

Since there is no information given about the thermos bottle, you can only make an approximation. In this case you might want to assume that the mixing and heat exchange of the two water samples is fast compared to the heat transfer to the thermos bottle, which is designed for good insulation, i.e. low heat transfer. Therefore, the heat transfer from the water to the thermos bottle may be neglected for the calculation.

Furthermore, you need to make an assumption about the pressure; however, the enthalpy of liquids is only slightly affected by differences of reasonable values of pressure. For simplicity’s sake, you might want to assume a constant pressure of $p=1\ \mathrm{bar}$.

Thus, the problem can be reduced to a simple enthalpy balance $$H_\text{mixture}=H_\text{cold}+H_\text{hot}$$

The enthalpy for the cold sample can be obtained from the specific enthalpy of liquid water at $p=1\ \mathrm{bar}$ and $T_\text{cold}=3.0\ \mathrm{^\circ C}$, which can be looked up in steam tables as $h_\text{cold}=0.012704\ \mathrm{kJ\ g^{-1}}$.

\begin{align}H_\text{cold}&=h_\text{cold}m_\text{cold}\\[6pt]&=0.012704\ \mathrm{kJ\ g^{-1}}\times125\ \mathrm g\\[6pt]&=15.88\ \mathrm{kJ}\end{align}

Likewise, the enthalpy for the hot sample can be obtained from the specific enthalpy of liquid water at $p=1\ \mathrm{bar}$ and $T_\text{hot}=91.0\ \mathrm{^\circ C}$, which is $h_\text{hot}=0.38127\ \mathrm{kJ\ g^{-1}}$.

\begin{align}H_\text{hot}&=h_\text{got}m_\text{hot}\\[6pt]&=0.38127\ \mathrm{kJ\ g^{-1}}\times283\ \mathrm g\\[6pt]&=107.90\ \mathrm{kJ}\end{align}

The resulting total enthalpy of the mixture is

\begin{align}H_\text{mixture}&=H_\text{cold}+H_\text{hot}\\[6pt]&=15.88\ \mathrm{kJ}+107.90\ \mathrm{kJ}\\[6pt]&=123.78\ \mathrm{kJ}\end{align}

The total mass of the mixture is

\begin{align}m_\text{mixture}&=m_\text{cold}+m_\text{hot}\\[6pt]&=125\ \mathrm g+283\ \mathrm g\\[6pt]&=408\ \mathrm g\end{align}

Therefore, the specific enthalpy of the mixture is

\begin{align}h_\text{mixture}&=\frac{H_\text{mixture}}{m_\text{mixture}}\\[6pt]&=\frac{123.78\ \mathrm{kJ}}{408\ \mathrm g}\\[6pt]&=0.30338\ \mathrm{kJ\ g^{-1}}\end{align}

Looking up the temperature for $h_\text{mixture}=0.30338\ \mathrm{kJ\ g^{-1}}$ at $p=1\ \mathrm{bar}$ shows $T_\text{mixture}=72.4\ \mathrm{^\circ C}$.

By way of comparison, you might even want to simplify further and assume that the specific heat capacity $c_p$ of water is constant so that the temperature of the mixture is given by

\begin{align}T_\text{mixture}&=\frac{T_\text{cold}m_\text{cold}+T_\text{hot}m_\text{hot}}{m_\text{cold}+m_\text{hot}}\\[6pt]&=\frac{3.0\ \mathrm{^\circ C}\times125\ \mathrm g+91.0\ \mathrm{^\circ C}\times283\ \mathrm g}{125\ \mathrm g+283\ \mathrm g}\\[6pt]&=64\ \mathrm{^\circ C}\end{align}