Question
Question: Define coefficient of thermal conductivity of glass \[2.2 \times {10^{ - 3}}cals/cm/kg\].Calculate t...
Define coefficient of thermal conductivity of glass 2.2×10−3cals/cm/kg.Calculate the rate of loss of heat through a glass window of area 1000cm2 and thickness 0.4cm when temperature inside is 37∘C and outside is−5∘C.
Solution
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 ΔtΔQ 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∘C on the inside and exposed to a temperature of −5∘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 ,
ΔtΔQ=ΔwKAΔT
Where, Kis coefficient of thermal conductivity
Ais the cross-sectional area of the material
Δwis the change in width of the material
ΔTis the change in temperature.
In our question two temperatures T1and T2are given, where T1 is temperature inside the glass and T2is temperature outside the glass.
T1=273+37=310K
T2=273−5=268K
Substituting the values on the above mentioned formula we get,
⇒ΔtΔQ=0.42.2×10−3×1000×(310−268)
⇒ΔtΔQ=0.42.2×10−3×1000×(42)
∴ΔtΔQ=231cal/s
Therefore, 231cal/sof 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.