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 $P/\rho$, where Pis the difference between the applied pressure and the pressure where the chocolate emerges from the tube, and 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$150\text{kJ/kg}$ . Assume that all of the work goes into that melting and that these fats make up $30\%$of the chocolates mass. What percentage of the fats melt during the extrusion if $p=5.5\text{MPaand}r={\text{1200kg/m}}^{\text{3}}$?

Short answer

The percentage of melting fats during the extrusion is $10\%$

Step-by-step solution

Identification of given data

i) The heat of fusion of fat is $L_f=150\frac{\text{kJ}}{\text{kg}}\text{or150}\times {\text{10}}^{\text{3}}\text{J/kg}$

ii) The difference between the applied pressure and the pressure where the chocolate emerges from the tube is $P=5.5\text{MPaor}5.5\times {10}^6\text{Pa}$

iii) The density of the chocolate is $\rho =1200{\text{kg/m}}^{\text{3}}$

iv) The work done per unit mass by the body, $\frac{W}{M}=\frac{P}{\rho }$

Understanding the concept

Upon the pressure applied on the chocolate to move it up to a given distance, some amount of the chocolate gets reduced due to the heat released in the force applied process. Applying constant pressure on the chocolate, the amount of heat released is given by its heat of fusion which is heat released when the chocolate melts its body to pass through the tube. We can find the equation for the work done per unit melting mass of the chocolate. Then, using the concept of the heat of fusion, we can find the percentage of melting fats during the extrusion.

Formulae:

The heat released in the process by the body, $Q=L_{f}m$ …(i)

Where, $L_f$is the latent heat of fusion, $m$ is the mass of the substance.

Determining the percentage of melting fats during the extrusion

The total mass of the chocolate is$M$and the mass of the melting chocolate is.

The work per unit melting mass of the chocolate is given, hence the work done by the body can be given as:

$W=\frac{P}{\rho }M$ …(ii)

There is phase change from the solid to liquid, and hence, the heat of transformation involved in this process is called heat of fusion and is given using equation (i).

For the work of extrusion of chocolate which takes place due to the energy can given the mass of the melting chocolates as:

$\begin{array}{c}Q=W\text{(Workdoneisequaltoheatoffusion)}\\ L_{f}m=\frac{P}{\rho }M\text{(fromequations(ii)and(i))}\\ m=\frac{PM}{\rho L_f}\\ =\frac{5.5\times {10}^6\text{Pa}}{1200{\text{kg/m}}^{\text{3}}\times 150\times {10}^3\text{J/kg}}M\\ =0.0306M\end{array}$

According to the problem, the mass of the fats is equal toof the total mass. Hence, the required mass of chocolates is given by

The percentage of melting fats during the extrusion is given as:

$\begin{array}{c}Percentage=\frac{0.0306M}{0.30M}\times 100\\ =10.2\%\\ \approx 10\%\end{array}$

Hence, the required value of melting percentage is$10\%$