
Define coefficient of thermal conductivity of glass \[2.2 \times {10^{ - 3}}cals/cm/kg\].Calculate the rate of loss of heat through a glass window of area \[1000c{m^2}\] and thickness \[0.4cm\] when temperature inside is \[{37^ \circ }C\] and outside is\[ - {5^ \circ }C\].
Answer
154.8k+ views
Hint The heat flow rate of material with area A and thickness w depends on the thermal conductivity of the material. In our statement , Heat flows from the higher temperature to the lower temperature. Use \[\dfrac{{\Delta Q}}{{\Delta t}}\] to find the rate of heat loss.
Complete Step By Step Solution
Coefficient of thermal conductivity is defined as the parameter of the material whose rate at which heat is transferred via conduction through the whole cross sectional area of the area. The coefficient of thermal conductivity is given by the alphabet K.
In our given question, it is given that there’s a glass material subjected to a heat of \[{37^ \circ }C\] on the inside and exposed to a temperature of \[ - {5^ \circ }C\] on the outside. The area and the change in thickness is given and asked on the rate of heat loss. Rate of change of heat can be calculated by using the formula ,
\[\dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{KA\Delta T}}{{\Delta w}}\]
Where, \[K\]is coefficient of thermal conductivity
\[A\]is the cross-sectional area of the material
\[\Delta w\]is the change in width of the material
\[\Delta T\]is the change in temperature.
In our question two temperatures \[{T_1}\]and \[{T_2}\]are given, where \[{T_1}\] is temperature inside the glass and \[{T_2}\]is temperature outside the glass.
\[{T_1} = 273 + 37 = 310K\]
\[{T_2} = 273 - 5 = 268K\]
Substituting the values on the above mentioned formula we get,
\[ \Rightarrow \dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{2.2 \times {{10}^{ - 3}} \times 1000 \times (310 - 268)}}{{0.4}}\]
\[ \Rightarrow \dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{2.2 \times {{10}^{ - 3}} \times 1000 \times (42)}}{{0.4}}\]
\[ \therefore \dfrac{{\Delta Q}}{{\Delta t}} = 231cal/s\]
Therefore, \[231cal/s\]of heat is lost during the following process in a glass material.
Note Rate of heat transfer is defined as the amount of heat that is transferred from one end to another along its cross sectional area per unit time. It is generally calculated in watts. Rate of heat loss is generally calculated by rate of change of heat since there is some loss from initial to final value.
Complete Step By Step Solution
Coefficient of thermal conductivity is defined as the parameter of the material whose rate at which heat is transferred via conduction through the whole cross sectional area of the area. The coefficient of thermal conductivity is given by the alphabet K.
In our given question, it is given that there’s a glass material subjected to a heat of \[{37^ \circ }C\] on the inside and exposed to a temperature of \[ - {5^ \circ }C\] on the outside. The area and the change in thickness is given and asked on the rate of heat loss. Rate of change of heat can be calculated by using the formula ,
\[\dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{KA\Delta T}}{{\Delta w}}\]
Where, \[K\]is coefficient of thermal conductivity
\[A\]is the cross-sectional area of the material
\[\Delta w\]is the change in width of the material
\[\Delta T\]is the change in temperature.
In our question two temperatures \[{T_1}\]and \[{T_2}\]are given, where \[{T_1}\] is temperature inside the glass and \[{T_2}\]is temperature outside the glass.
\[{T_1} = 273 + 37 = 310K\]
\[{T_2} = 273 - 5 = 268K\]
Substituting the values on the above mentioned formula we get,
\[ \Rightarrow \dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{2.2 \times {{10}^{ - 3}} \times 1000 \times (310 - 268)}}{{0.4}}\]
\[ \Rightarrow \dfrac{{\Delta Q}}{{\Delta t}} = \dfrac{{2.2 \times {{10}^{ - 3}} \times 1000 \times (42)}}{{0.4}}\]
\[ \therefore \dfrac{{\Delta Q}}{{\Delta t}} = 231cal/s\]
Therefore, \[231cal/s\]of heat is lost during the following process in a glass material.
Note Rate of heat transfer is defined as the amount of heat that is transferred from one end to another along its cross sectional area per unit time. It is generally calculated in watts. Rate of heat loss is generally calculated by rate of change of heat since there is some loss from initial to final value.
Recently Updated Pages
Young's Double Slit Experiment Step by Step Derivation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

Trending doubts
If a wire of resistance R is stretched to double of class 12 physics JEE_Main

Explain the construction and working of a GeigerMuller class 12 physics JEE_Main

Wheatstone Bridge for JEE Main Physics 2025

The force of interaction of two dipoles if the two class 12 physics JEE_Main

Three charges sqrt 2 mu C2sqrt 2 mu Cand sqrt 2 mu class 12 physics JEE_Main

The potential of A is 10V then the potential of B is class 12 physics JEE_Main

Other Pages
JEE Advanced 2025 Revision Notes for Mechanics

Ideal and Non-Ideal Solutions Raoult's Law - JEE

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

An uncharged sphere of metal is placed inside a charged class 12 physics JEE_Main

Three mediums of refractive indices mu 1mu 0 and mu class 12 physics JEE_Main

A signal of 5kHz frequency is amplitude modulated on class 12 physics JEE_Main
