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Chapter 28: Problem 9

Water at \(50^{\circ} \mathrm{C}\) is filled in a closed cylindrical vessel of height \(10 \mathrm{~cm}\) and cross sectional area \(10 \mathrm{~cm}^{2}\). The walls of the vessel are adiabatic but the flat parts are made of \(1-\mathrm{mm} \quad\) thick \(\quad\) aluminium \(\quad\left(K=200 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-1}{ }^{-1} \mathrm{C}^{-1}\right)\). Assume that the outside temperature is \(20^{\circ} \mathrm{C}\). The density of water is \(1000 \mathrm{~kg} \mathrm{~m}^{-5}\), and the specific heat capacity of water \(=4200 \mathrm{~J} \mathrm{k} \stackrel{-\mathrm{l}}{\mathrm{g}} \mathrm{C}^{-1}\). Estimate the time taken for the temperature to fall by \(1 \cdot 0^{\circ} \mathrm{C}\). Make any simplifying assumptions you need but specify them.

Short Answer

Expert verified
Time to cool by 1掳C is 0.035 seconds.

Step by step solution

01

Understand the Problem

This problem involves the heat loss through the flat parts of a cylindrical vessel to find the time required for the temperature to drop by 1掳C. The cylindrical vessel is adiabatic, except for its flat surfaces, made of aluminum. The problem provides values for thermal conductivity, specific heat, density, and the thickness of aluminum.
02

Identify Simplifying Assumptions

Assume that the temperature change occurs uniformly, and heat loss by conduction is the only mode of heat transfer. Additionally, neglect heat loss through the curved surface and treat the vessel as having constant properties over the small temperature change.
03

Calculate Surface Area for Heat Transfer

The flat parts of the cylinder are its top and bottom. Since the cross-sectional area of the vessel is given as \(10 \mathrm{\ cm}^2\), both top and bottom sum to a total area \(A = 2 \times 10 \mathrm{\ cm}^2 = 20 \mathrm{\ cm}^2 = 0.002 \mathrm{\ m}^2\).
04

Calculate the Mass of the Water

The volume of water is the product of its base area and height, yielding \(V = 10 \mathrm{\ cm}^2 \times 10 \mathrm{\ cm} = 100 \mathrm{\ cm}^3 = 0.0001 \mathrm{\ m}^3\). Given the density of water \(\rho = 1000 \mathrm{\ kg} / \mathrm{m}^3\), the mass \(m = \rho \times V = 1000 \times 0.0001 = 0.1 \mathrm{\ kg}\).
05

Determine Initial and Final Temperatures

The initial temperature is \(T_i = 50^{\circ} \mathrm{C}\) and the final temperature after cooling is \(T_f = 49^{\circ} \mathrm{C}\). Therefore, \(\Delta T = T_i - T_f = 1^{\circ} \mathrm{C}\).
06

Calculate the Heat Lost by Water

Using the formula for heat \(Q = mc\Delta T\), where \(m = 0.1 \mathrm{\ kg}\), \(c = 4200 \mathrm{\ J/kg^{\circ}C}\), and \(\Delta T = 1^{\circ} \mathrm{C}\), the heat lost \(Q = 0.1 \times 4200 \times 1 = 420 \mathrm{\ J}\).
07

Use Fourier's Law to Find Heat Conduction

The rate of heat conduction through the aluminum plates is given by Fourier's law: \(\frac{Q}{t} = \frac{KA(T_{inside} - T_{outside})}{d}\), where \(K = 200 \mathrm{\ J/s\ m\ ^\circ C}\), \(A = 0.002 \mathrm{\ m}^2\), \(d = 0.001 \mathrm{\ m}\), \(T_{inside} = 50^{\circ}C\), and \(T_{outside} = 20^{\circ}C\). Substitute the values to get \(\frac{Q}{t} = \frac{200 \times 0.002 \times (50-20)}{0.001} = 12000 \mathrm{\ J/s}\).
08

Solve for Time to Achieve Temperature Change

Rearranging the equation to find \(t\), we get \(t = \frac{Q}{\frac{Q}{t}} = \frac{420}{12000} = 0.035 \mathrm{\ s}\). So, the time taken for the temperature to fall by \(1^{\circ} \mathrm{C}\) is \(0.035 \mathrm{\ s}\).

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Key Concepts

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

Heat Conduction
Heat conduction is the process through which thermal energy is transferred from one material to another or within a material. This occurs when there is a temperature difference, causing energy to move from the hotter area to the cooler one.
One way to visualize this is by imagining a pot getting hot on a stove. The heat from the stove travels through the metal pot, finally making it hot enough to cook food. In our exercise, the aluminum parts of the cylindrical vessel allow heat to be conducted from the warm water to the cooler surroundings.
Heat conduction is measured using the material鈥檚 thermal conductivity, denoted as \(K\). The higher the \(K\), the more efficiently heat is conducted. In our example, aluminum has a high \(K\), meaning it鈥檚 excellent at conducting heat away from the water.
Specific Heat Capacity
Specific heat capacity is a concept that describes how much heat energy a substance can hold. Specifically, it鈥檚 the amount of heat needed to change the temperature of 1 kilogram of a material by 1 degree Celsius.
Think of it as a measure of a material's resistance to temperature change. Water, for instance, has a high specific heat capacity of 4200 \(J/kg^{\circ}C\), which is a reason it can retain heat for a long time.
In practical terms, this means if you heat water, it takes a lot of energy to raise its temperature. Conversely, it also takes a significant amount of time to cool down, hence influencing the time it takes for the water in the vessel to decrease in temperature.
Temperature Change
Temperature change refers to the difference between the initial and final temperatures of a substance as it gains or loses heat. In thermodynamic problems, this is often what you want to calculate or control.
Taking our exercise as a practical example, the initial temperature of water inside the vessel is \(50^{\circ}C\) and the goal is to reduce it to \(49^{\circ}C\), resulting in a \(\Delta T\) of \(1^{\circ}C\).
To calculate the energy exchanged due to this temperature change, the formula \(Q = mc\Delta T\) is used, where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.
Adiabatic Process
An adiabatic process is a thermodynamic process in which no heat is exchanged with the surroundings. This means that whatever happens within the system, the outside has no direct influence through heat transfer.
Adiabatic walls are perfect insulators, and they don鈥檛 allow any thermal exchange. In our exercise, the side walls of the cylindrical vessel are adiabatic, ensuring that heat transfer can only happen through the top and bottom aluminum sections.
Understanding that only specific areas allow heat conduction is crucial for correctly setting up the calculation of how long it takes for the temperature to change by one degree.

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Most popular questions from this chapter

A room has a window fitted with a single \(1-0 \mathrm{~m} \times 2 \cdot 0 \mathrm{~m}\) glass of thickness \(2 \mathrm{~mm}\). (a) Calculate the rate of heat flow through the closed window when the temperature inside the room is \(32^{\circ} \mathrm{C}\) and that outside is \(40^{\circ} \mathrm{C} .\) (b) The glass is now replaced by two glasspanes, each having a thickness of \(1 \mathrm{~mm}\) and separated by a distance of \(1 \mathrm{~mm}\). Calculate the rate of heat flow under the same conditions of temperature. Thermal conductivity of window glass \(=1 \cdot 0 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\) and that of air \(=0 \cdot 025 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\).

On a winter day when the atmospheric temperature drops to \(-10^{\circ} \mathrm{C}\), ice forms on the surface of a lake. (a) Calculate the rate of increase of thickness of the ice when \(10 \mathrm{~cm}\) of ice is already formed. (b) Calculate the total time taken in forming \(10 \mathrm{~cm}\) of ice. Assume that the temperature of the entire water reaches \(0^{\circ} \mathrm{C}\) before the ice starts forming. Density of water \(=1000 \mathrm{~kg} \mathrm{~m}^{-3}\), latent heat of fusion of ice \(=3.36 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}\) and thermal conductivity of ice \(=1.7 \mathrm{~W} \mathrm{~m}^{-1}{ }^{\circ} \mathrm{C}^{-1}\). Neglect the expansion of water on freezing.

A spherical ball \(A\) of surface area \(20 \mathrm{~cm}^{2}\) is kept at the centre of a hollow spherical shell \(B\) of area \(80 \mathrm{~cm}^{2}\). The surface of \(A\) and the inner surface of \(B\) emit as blackbodies. Both \(A\) and \(B\) are at \(300 \mathrm{~K}\). (a) How much is the radiation energy emitted per second by the ball \(A ?\) (b) How much is the radiation energy emitted per second by the inner surface of \(B\) ? (c) How much of the energy emitted by the inner surface of \(B\) falls back on this surface itself ?

A spherical tungsten piece of radius \(1 \cdot 0 \mathrm{~cm}\) is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The piece is maintained at \(1000 \mathrm{~K}\) by heating it electrically. Find the rate at which the electrical energy must be supplied. The emissivity of tungsten is \(0: 30\) and the Stefan constant \(\sigma\) is \(6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}\)

A hot body placed in a surrounding of temperature \(\theta_{0}\) obeys Newton's law of cooling \(\frac{d \theta}{d t}=-k\left(\theta-\theta_{0}\right) .\) Its temperature at \(t=0\) is \(\theta_{1} .\) The specific heat capacity of the body is \(s\) and its mass is \(m\). Find (a) the maximum heat that the body can lose and (b) the time starting from \(t=0\) in which it will lose \(90 \%\) of this maximum heat.
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