91Ó°ÊÓ

ecp A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique is to measure the temperature difference between the input and output points of a flowing stream of the liquid while adding energy at a known rate. (a) Start with the equations \(Q=m c(\Delta T)\) and \(m=\rho V\), and show that the rate at which energy is added to the liquid is given by the expression \(\Delta Q / \Delta \mathrm{t}=\rho c(\Delta T)(\Delta V / \Delta t)\), (b) In a particular experiment, a liquid of density \(0.72 \mathrm{~g} / \mathrm{cm}^{3}\) flows through the calorimcter at the rate of \(3.5 \mathrm{~cm}^{3} / \mathrm{s}\). At steady state, a temperature difference of \(5.8^{\circ} \mathrm{C}\) is established between the input and output points when energy is supplied at the rate of \(40 \mathrm{~J} / \mathrm{s}\). What is the specific heat of the liquid?

Step by step solution

01

Derive the Equation

First, let's substitute \(m = \rho V\) into the \(Q = mc \Delta T\) equation to get: \(Q = \rho V c \Delta T\) . To get how much energy is being added per unit time (\(\Delta Q / \Delta t\)), divide both sides of the equation by \(\Delta t\). This results in \(\Delta Q / \Delta t = \rho c \Delta T (V / \Delta t)\). Here \(\Delta V / \Delta t\) is the rate at which the volume of the liquid is flowing. We now have \(\Delta Q / \Delta t = \rho c \Delta T (\Delta V / \Delta t)\) as required.

02

Substitute Values into Equation

Given that the liquid is flowing at a rate of \(3.5 \,cm^{3}/s\), this is \( \Delta V / \Delta t\). The difference in temperature (\(\Delta T\)) is given as \(5.8^{\circ}C\), and \(\rho\) the density of the liquid is given as \(0.72 g/cm^{3}\) (but we need to convert this to \( Kg/m^{3}\) for proper units), and finally, \(\Delta Q / \Delta t\) is given as \(40 J/s\). So substituting these values in, gives us the equation: \(40 = 72000c(5.8)(3.5 \times 10^{-6})\).

03

Solve for c

Rearrange the equation to solve for \(c\), the specific heat capacity. Doing this yields the equation for \(c\) as: \(c = 40 / [72000 \cdot 5.8 \cdot 3.5 \times 10^{-6}]\). After doing the calculation, \(c \) will be approximately \(244.76 J/(Kg \cdot C)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Flow Calorimeter

Understanding the concept of a flow calorimeter is essential for any student studying the physical properties of liquids. A flow calorimeter is a device used in thermodynamics to determine the specific heat capacity of a fluid. It operates by measuring the change in temperature of a liquid as it passes from the inlet point, receives a certain amount of energy, and reaches the outlet.

This apparatus is invaluable in experimental settings because it can provide real-time data on how a liquid absorbs and dissipates thermal energy while in motion. Unlike in static calorimetry, where the liquid is contained and its temperature change is observed, the flow calorimeter deals with fluid dynamics and continuous energy input, thereby offering insights into the liquid's behavior under conditions closer to real-world applications.

Thermal Energy Transfer

Thermal energy transfer refers to the movement of heat from one place to another. In the context of a flow calorimeter, this transfer is measured as the liquid absorbs thermal energy when energy is added at a known rate. The resulting temperature increase of the liquid is monitored, providing data needed to calculate the specific heat capacity.

It's important to understand that thermal energy transfer is governed by principles of thermodynamics, particularly the first law, which states that energy is conserved. In the calorimeter, the energy supplied to the liquid is either stored as increased internal energy or transferred elsewhere, resulting in a measurable temperature change.

Thermodynamics

Thermodynamics is a fundamental concept in physics that studies the effects of temperature, pressure, and volume on physical systems at the macroscopic scale. It lays out the principles governing thermal energy transfer, which are essential to understanding flow calorimeter experiments.

For instance, when using the equation \(Q = m c \Delta T\), we make use of the first law of thermodynamics, which, as previously explained, implies energy conservation in the system. Indeed, being comfortable with the laws of thermodynamics allows students to accurately interpret the energy transfers taking place within the flow calorimeter and other thermal systems.

Physical Properties of Liquids

Physical properties of liquids, such as specific heat capacity, density, and viscosity, play a crucial role in experiments using a flow calorimeter. The specific heat capacity is a measure of how much thermal energy a substance can hold, which directly influences how its temperature changes when subjected to energy transfer.

Knowing the density of the liquid (\( \rho \)) is also vital as it factors into the calculation for the rate of energy addition, \( \Delta Q / \Delta t = \rho c \Delta T (\Delta V / \Delta t) \). Each physical property contributes to the detailed analysis of the liquid's behavior under various thermal conditions, which illustrates why it is important for students to grasp the relevance of these physical characteristics.