Question

In the extrusion of cold chocolate from a tube, work is done on the chocolate by the pressure applied by a ram forcing the chocolate through the tube. The work per unit mass of extruded chocolate is equal to $\dfrac{p}{\rho }$, where $p$ is the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and $\rho$ is the density of the chocolate. Rather than increasing the temperature of the chocolate, this work melts cocoa fats in the chocolate. These fats have a heat of fusion of $150KJk{{g}^{-1}}$. Assume that all of the work goes into that melting and that these fats make up $30$% of the chocolate’s mass. What percentage of the fats make during the extrusion if $p=5.5MPa$ and $\rho =1200kg{{m}^{-3}}$?

Answer 1

Work done by the application of pressure by a ram is equal to the heat required for the phase transition of solid chocolate to liquid chocolate. Heat required for phase transition from solid phase to liquid phase is equal to the product of latent heat of fusion and mass of the substance which undergoes phase transition.
Formula used:
$1)\dfrac{W}{{{M}_{total}}}=\dfrac{p}{\rho }$
$2)W=Q$
$3)Q=m{{L}_{f}}$

Complete answer:
Here, we are dealing with the phase transition of solid chocolate to liquid chocolate. Rather than increasing the temperature, phase transition occurring in this case is enhanced with the help of work done by the application of pressure on the tube, which contains solid chocolate. We are told that work done per unit mass of extruded chocolate is equal to $\dfrac{p}{\rho }$, where $p$ is the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and $\rho$ is the density of the chocolate. Clearly,
$\dfrac{W}{{{M}_{total}}}=\dfrac{p}{\rho }$
where
$W$ is the work done by the application of pressure on the tube containing solid chocolate
${{M}_{total}}$ is the total mass of the chocolate which undergoes phase transition
Let this be equation 1.
Now, we are told that all the work done by the application of pressure melts cocoa fats in the chocolate. This suggests that work done by the application of pressure is equal to the heat required for the melting process of cocoa fats in the chocolate. Clearly, we have
$W=Q$
where
$W$ is the work done by the application of pressure on the tube containing solid chocolate
$Q$ is the heat required for the melting process of cocoa fats in the chocolate
Let this be equation 2.
Combining equation 1 and equation 2, we have
$\dfrac{W}{{{M}_{total}}}=\dfrac{p}{\rho }\Rightarrow \dfrac{Q}{{{M}_{total}}}=\dfrac{p}{\rho }$
Let this be equation 3.
Now, we know that heat required for melting process of a solid is given by
$Q=m{{L}_{f}}$
where
$Q$ is the heat required for the melting process of a solid substance
$m$ is the mass of solid substance
${{L}_{f}}$ is the latent heat of fusion of the substance
Let this be equation 4.
Substituting equation 4 in equation 3, we have
$\dfrac{m{{L}_{f}}}{{{M}_{total}}}=\dfrac{p}{\rho }\Rightarrow m=\dfrac{p{{M}_{total}}}{\rho {{L}_{f}}}$
where
$m$ is the mass of cocoa fats in the chocolate which undergo melting
${{L}_{f}}$ is the latent heat of fusion of cocoa fats in the chocolate
${{M}_{total}}$ is the total mass of the chocolate which undergoes phase transition
$p$ is the difference between the applied pressure and the pressure where the chocolate emerges from the tube
$\rho$ is the density of the chocolate.
Let this be equation 5.
Now, from the question, we are given that the total mass of cocoa fats in the chocolate is equal to $30$% of the total mass of chocolate. Clearly, if ${{M}_{fat}}$ represents the total mass of cocoa fats in the chocolate, then, ${{M}_{fat}}$ is given by
${{M}_{fat}}=0.3{{M}_{total}}$
where
${{M}_{fat}}$ is the total mass of cocoa fats in the chocolate
${{M}_{total}}$ is the total mass of chocolate
Let this be equation 6.
Dividing equation 5 by equation 6, we have
$\dfrac{m}{{{M}_{fat}}}=\dfrac{p{{M}_{total}}}{\rho {{L}_{f}}(0.3{{M}_{total}})}=\dfrac{p}{\rho {{L}_{f}}\times 0.3}$
Let this be equation 7.
From the question, we know that
$\begin{aligned} & p=5.5MPa \\\ & \rho =1200kg{{m}^{-3}} \\\ & {{L}_{f}}=150KJk{{g}^{-1}} \\\ \end{aligned}$
Substituting these values in equation 7, we have
$\dfrac{m}{{{M}_{fat}}}=\dfrac{p}{\rho {{L}_{f}}\times 0.3}=\dfrac{5.5MPa}{1200kg{{m}^{-3}}\times 150KJk{{g}^{-1}}\times 0.3}=\dfrac{5.5\times {{10}^{6}}Pa}{1200kg{{m}^{-3}}\times 150\times {{10}^{3}}Jk{{g}^{-1}}\times 0.3}=0.101$
Taking percentage of the same, we have
$\dfrac{m}{{{M}_{fat}}}\times 100$%$=0.101\times$ $100$%$=10.1$%
Therefore, the percentage of cocoa fats which undergo melting when the given pressure is applied on the tube containing the chocolate is equal to $10.1$%.

Note:
Students need to understand the logic of the question properly before starting to attempt the question. Deducing equation 1, equation 6 and equation 7 are the three main sections of the solution, which need careful logical thinking. Also, students need to be thorough with conversion formulas. Conversion formulas used in this solution are:
$1MPa={{10}^{6}}Pa$
$1KJ={{10}^{3}}J$
$1Pa=1J{{m}^{-3}}$