
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
219k+ 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
A square frame of side 10 cm and a long straight wire class 12 physics JEE_Main

The work done in slowly moving an electron of charge class 12 physics JEE_Main

Two identical charged spheres suspended from a common class 12 physics JEE_Main

According to Bohrs theory the timeaveraged magnetic class 12 physics JEE_Main

ill in the blanks Pure tungsten has A Low resistivity class 12 physics JEE_Main

The value of the resistor RS needed in the DC voltage class 12 physics JEE_Main

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Understanding Uniform Acceleration in Physics

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Understanding Atomic Structure for Beginners

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

