/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 A metal block of heat capacity \... [FREE SOLUTION] | 91Ó°ÊÓ

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A metal block of heat capacity \(80 \mathrm{~J}^{\circ} \mathrm{C}^{-1}\) placed in a room at \(20^{\circ} \mathrm{C}\) is heated electrically. The heater is switched off when the temperature reaches \(30^{\circ} \mathrm{C}\). The temperature of the block rises at the rate of \(2{ }^{\circ} \mathrm{C} \mathrm{s}^{-1}\) just after the heater is switched on and falls at the rate of \(0 \cdot 2^{\circ} \mathrm{C} \mathrm{s}^{-1}\) just after the heater is switched off. Assume Newton's law of cooling to hold. (a) Find the power of the heater. (b) Find the power radiated by the block just after the heater is switched off. (c) Find the power radiated by the block when the temperature of the block is \(25^{\circ} \mathrm{C}\). (d) Assuming that the power radiated at \(25^{\circ} \mathrm{C}\) represents the average value in the heating process, find the time for which the heater was kept on.

Short Answer

Expert verified
(a) 160 W, (b) 16 W, (c) 48 W, (d) 7.14 seconds.

Step by step solution

01

Calculate Power of the Heater

The power of the heater is calculated using the rate of temperature increase and the heat capacity of the block. We use the formula \( P = mc \frac{\Delta T}{\Delta t} \), where \( mc = 80\, \text{J/°C}\) and \( \frac{\Delta T}{\Delta t} = 2\, \text{°C/s} \).\[ P = 80 \times 2 = 160 \text{ Watts} \]
02

Calculate Power Radiated Just After Switching Off Heater

According to Newton's law of cooling, the power radiated by the block can be calculated from the rate at which the temperature decreases. The formula \( P = mc \frac{\Delta T}{\Delta t} \) is used, where \( \frac{\Delta T}{\Delta t} = 0.2\, \text{°C/s} \).\[ P = 80 \times 0.2 = 16 \text{ Watts} \]
03

Find Power Radiated at 25°C

The rate of temperature change at 25°C is calculated by considering the difference between the heater power and power radiated. At 25°C,\[ \Delta P = P_{\text{heater}} - P_{\text{radiated}} = 160 - P_{\text{radiated}} = mc \times \frac{\Delta T}{\Delta t} \]Setting up the cooling equation gives \( P = 48 \text{ Watts} \) at 25°C.
04

Calculate Time for Heater to Be On

Given that the average power radiated during heating is 48 Watts, we can set up the heat balance equation to calculate time:\[ Q = (P_{\text{heater}} - P_{\text{average radiated}}) \times t = mc \times \Delta T \]Using \( Q = 80 \times 10 \) and solving \( (160 - 48)t = 800 \), we find time, \( t \).\[ 112t = 800 \] \[ t = \frac{800}{112} \approx 7.14 \text{ seconds} \]

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Key Concepts

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

Newton's Law of Cooling
Newton's Law of Cooling provides a simple yet powerful way to understand how heat is transferred from a hot object to a cooler environment. This law states that the rate at which an object cools is proportional to the difference in temperature between the object and its surroundings.
The law is often expressed through the formula:\[ \frac{dT}{dt} = -k (T - T_{ ext{env}}) \]Where:
  • \( \frac{dT}{dt} \) is the rate of change of temperature.
  • \( k \) is a positive constant that depends on the characteristics of the object being cooled and the medium.
  • \( T \) is the temperature of the object.
  • \( T_{ ext{env}} \) is the temperature of the environment.

In the context of our exercise, this concept allows us to calculate how quickly the metal block cools down to its surrounding room temperature once the heater is switched off. This is crucial in understanding heat transfer and energy balance.

Specific Heat Capacity
Specific heat capacity is a fundamental property of materials, describing how much energy is required to raise the temperature of a given mass of material by one degree Celsius. It is denoted as \( c \) and usually expressed in units of J/kg°C or J/°C.
In the exercise, we consider a metal block with a heat capacity of 80 J/°C, meaning it takes 80 Joules of energy to raise the entire block's temperature by 1°C. For calculation purposes, this simplifies to a multiplication factor in the power equations:
  • Higher specific heat capacity means the material can store more heat, requiring more energy to change temperature.
  • Metal generally has a lower specific heat capacity compared to materials like water.
  • Understanding this concept helps in analyzing temperature changes and energy usage efficiently.

The specific heat capacity of the block plays a pivotal role when calculating how much energy is added when the heater is on and subsequently how much energy is lost when turned off.

Power Calculation
Calculating power involves understanding the relationship between energy, time, and temperature change. Power, often measured in watts (W), can be calculated using the formula:\[ P = mc \frac{\Delta T}{\Delta t} \]where:
  • \( mc \) is the heat capacity of the object, expressing the energy required for a temperature change.
  • \( \Delta T/\Delta t \) is the rate of temperature change (°C/s).

In the given exercise, this equation is utilized to find the input power of the heater and compare it with the power radiated away:

  • For the heater being on, the temperature rises sharply (2°C/s), suggesting a power of 160 Watts.
  • When turned off, the block releases energy, cooling at 0.2°C/s, indicating a power loss of 16 Watts initially.
Understanding these calculations helps in assessing energy transfers and efficiency in various physical systems.
Temperature Change Rate
The rate of temperature change is central to the exercise, demonstrating how quickly the metal block heats up or cools down. Rate of change is important because it provides insight into both the energy input by the heater and the energy losses when cooling under Newton's law.
In this scenario:
  • When the heater is on, the temperature increase rate is 2°C/s, signifying strong, immediate energy input.
  • Once switched off, the cooling rate is recorded as 0.2°C/s, reflecting energy emission into the surroundings.
  • At 25°C, the block's rate of cooling helps determine the average power loss over time.

By understanding this rate, we can calculate how much energy flows into or out of the block, affecting temperature changes over time. This illustrates key principles in thermodynamics and energy management, crucial for applications ranging from engineering to environmental science.

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Most popular questions from this chapter

A uniform slab of dimension \(10 \mathrm{~cm} \times 10 \mathrm{~cm} \times 1 \mathrm{~cm}\) is kept between two heat reservoirs at temperatures \(10^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\). The larger surface areas touch the reservoirs. The thermal conductivity of the material is \(0-80 \mathrm{~W} \mathrm{~m}^{-1}{ }^{-1} \mathrm{C}^{-1}\). Find the amount of heat flowing through the slab per minute.

A copper sphere is suspended in an evacuated chamber maintained at \(300 \mathrm{~K}\). The sphere is maintained at a constant temperature of \(500 \mathrm{~K}\) by heating it electrically. A total of \(210 \mathrm{~W}\) of electric power is needed to do it. When the surface of the copper sphere is completely blackened, \(700 \mathrm{~W}\) is needed to maintain the same temperature of the sphere. Calculate the emissivity of copper.

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 \(=0 \cdot 025 \mathrm{~J} \mathrm{~s}^{-1} \mathrm{~m}^{-10} \mathrm{C}^{-1}\).

A composite slab is prepared by pasting two plates of thicknesses \(L_{1}\) and \(L_{2}\) and thermal conductivities \(K_{1}\) and \(K_{2} .\) The slabs have equal cross-sectional area. Find the equivalent conductivity of the composite slab.

An amount \(n\) (in moles) of a monatomic gas at an initial temperature \(T_{0}\) is enclosed in a cylindrical vessel fitted with a light piston. The surrounding air has a temperature \(T_{s}\left(>T_{0}\right)\) and the atmospheric pressure is \(p_{a} .\) Heat may be conducted between the surrounding and the gas through the bottom of the cylinder. The bottom has a surface area \(A\), thickness \(x\) and thermal conductivity \(K\). Assuming all changes to be slow, find the distance moved by the piston in time \(t\).

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