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Chapter 12: Problem 34

The rales of fall o temperatures of two identical solid spheres of different materials are equal at a certain temperature (a) their specific heat capacities are equal (b) their heat capacitics are cqual (c) their specific heat capacities are proportional to their densities (d) their specific heat capacities are inversely proportional to their densities

Short Answer

Expert verified
(a) their specific heat capacities are equal.

Step by step solution

01

Understanding the Problem

We are given two identical solid spheres made of different materials that are cooling down at the same rate at a certain temperature. The task is to determine which of the given options regarding their specific heat capacities is true.
02

Recall Relevant Physics Concepts

The cooling rate of an object is described by Newton鈥檚 Law of Cooling, which involves the heat transfer properties of the material. The heat capacity of a material is defined as the amount of heat needed to change the temperature of an entire object by a unit degree, while the specific heat capacity is the heat capacity per unit mass.
03

Establish Equality of Heat Loss

When two objects cool at the same rate, their heat loss per unit time is the same. The heat loss rate can be equated for both spheres. For a sphere, the heat loss rate is proportional to the product of the mass, the specific heat capacity, and the rate of change of temperature.
04

Equate Heat Loss Expressions

Let the mass of each sphere be \( m \), specific heat capacities be \( c_1 \) and \( c_2 \), and densities be \( \rho_1 \) and \( \rho_2 \). Since the spheres are identical, their mass is the same, and thus the product \( m \, c_1 \frac{dT}{dt} = m \, c_2 \frac{dT}{dt} \), leading to \( c_1 = c_2 \).
05

Analyze Given Options

From the heat loss equality, \( c_1 = c_2 \) directly means their specific heat capacities are equal, which corresponds to option (a). Options involving relations to densities, options (c) and (d), or heat capacities, option (b), do not follow from the equality derived.

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Key Concepts

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

Specific Heat Capacity and Its Importance
The concept of specific heat capacity is fundamental in thermal physics. It measures how much heat energy is required to increase the temperature of a unit mass of a substance by one degree Celsius. A material with a high specific heat capacity can absorb a lot of heat without a significant change in temperature.
This property is crucial in understanding how different materials react to heat.
There are some key points to remember:
  • Water, for instance, has a high specific heat capacity, which is why it is effective in regulating temperature.
  • In comparison, metals generally have a lower specific heat capacity, meaning they heat up and cool down quicker.
Understanding specific heat capacity helps us analyze and predict how different materials respond when exposed to heat energy, which is essential for solving problems involving heat transfer and temperature changes.
Understanding Newton's Law of Cooling
Newton's Law of Cooling describes the rate at which an exposed body changes temperature through radiation. It states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature.
In simpler terms, the greater the difference, the faster the temperature change. Here鈥檚 a breakdown of the key aspects:
  • The law assumes the cooling occurs in a medium (like air) and that the medium's temperature remains constant.
  • This is an exponential relationship, meaning the rate of cooling decreases over time as the temperature difference reduces.
Newton's Law of Cooling is particularly useful in calculating how quickly objects cool and is applied in many thermal physics problems, including determining heat transfer rates in various materials.
Exploring the World of Thermal Physics
Thermal physics is a branch of physics that deals with heat and temperature and their relation to energy and work. It combines concepts from thermodynamics, statistical mechanics, and kinetic theory to understand physical processes where heat plays a significant role.
  • Thermal physics helps explain phenomena like heat conduction, convection, and radiation.
  • It also involves analyzing systems at thermal equilibrium鈥攚hen two objects in thermal contact do not exchange heat.
Learning about thermal physics enhances our ability to understand and predict how various materials interact with heat, which is vital across many scientific and engineering disciplines.
The Concept of Heat Capacity
Heat capacity refers to the amount of heat needed to raise the temperature of an entire object by one degree Celsius, which is different from specific heat capacity that focuses on a unit mass. This means that heat capacity is dependent on the mass and nature of the substance.
It is a macroscopic property and varies between substances and their forms (solid, liquid, gas).
  • For example, a large body of water has a significantly higher heat capacity compared to a small quantity, even though they have the same specific heat capacity.
  • Heat capacity is a crucial concept in defining system energy changes, particularly in large bodies of matter such as lakes or reactors.
Understanding heat capacity enables us to predict how much energy is needed to heat different-sized objects, important for practical applications in energy management and environmental science.

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Most popular questions from this chapter

A pitcher with \(1 \mathrm{~m}\) thick porous walls contains \(10 \mathrm{~kg}\) of water. Water comes to its outer surface and cvaporates at the rate of \(0.1 \mathrm{~g} / \mathrm{s}\). The surface arca of the pitcher (onc sidc) \(=200 \mathrm{~cm}^{2}\). The room temperature \(=42{ }^{\circ} \mathrm{C}\), latent heat of vaporization \(=2.27 \times 10^{6} \mathrm{~J} / \mathrm{kg}\) and the thermal conductivity of the porous walls \(=0.80 \mathrm{~J} / \mathrm{m}-\mathrm{s}-{ }^{\circ} \mathrm{C}\). Calculate the temperature of water in the piteher when il attains a constant value.

A highly conducting hollow cylinder of radius \(a\) and length \(L\) is surrounded by a co-axial layer of a material having thertnal conductivity \(K\) and negligible heat copacity, Temperature of surrounding space (out side the layer) is \(\theta_{2}\), which is higher than temperature of the cylinder. If heat capacity per unit volume of cylinder material is \(s\) and outer radius of tho layer is \(b\), calculate the thermal resistance of the cylindrical shell and also calculate the variation o " lemperaturc inside it with respect to the radial distance from the axis. Assumc cnd faces to be thermally insulaled.

The amount of heal energy radialed per second by a surface depends upon (a) area of the surface (b) dirrcrence of lemperalure bctween the surface and its surroundings (c) nalure of surface (d) nonc

Two bodies of same size \(P\) and \(Q\) are placed in a room which is maintained at a very low temperature. The bodies \(P\) and \(Q\) have temperatures \(527^{\circ} \mathrm{C}\) and \(127^{\circ} \mathrm{C}\) respectively. The energy radiated by \(P\) is (a) equal to \(Q\) (b) less than \(Q\) (c) greater than \(Q\) (d) 16 times greater than \(Q\)

Assuming no heat losses, the hoat released by the condensation of \(x \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\) can be used to convert \(y \mathrm{~g}\) of ico at \(0^{\circ} \mathrm{C}\) into water at \(100^{\circ} \mathrm{C}\), the ratio \(x: \mathrm{y}\) is (a) \(1: 1\) (b) \(1: 2\) (c) \(1: 3\) (d) \(3: 1\)
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