Answer
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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.
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