We want to measure the thermal conductivity of an unknown insulator material. Fo
ID: 2008194 • Letter: W
Question
We want to measure the thermal conductivity of an unknown insulator material. For this we use the following set-up: A 4mm thick plate of the unknown material is placed between two iron plates of thickness 3cm each. All three plates are 15cm by 15cm in size. The upper iron plate is heated to 350K and the lower iron plate is kept at 290K. Once a stationary temperature profile has developed across the insulator, the heater is removed from the upper iron plate. We observe that the temperature of the upper iron plate drops by 2.5K after 100 seconds. neglecting any loss of heat to the environment, what is the thermal conductivity coefficent for the unknown insulator material? Hint: the density of iron is p=7.9gcm3 and its specific heat capacity is 448 j/kg°C.
Explanation / Answer
Data:
Thickness of unknown material, L = 4 mm
= 4 x 10^-3 m
Temperature of upper iron plate, 1 = 350 K
Temperature of lower iron plate, 2 = 290 K
Drop in temperature of iron plate, = 2.5 K
Time, t = 100 s
Density of iron, = 7.9 g/cc
= 7900 kg/m^3
Specific heat of iron, c = 448 J/kg.K
Cross-sectional area of unknown material, A = 15 * 15
= 225 cm^2
= 225 x 10^-4 m^2
Volume of iron plate, V = 15 * 15 * 3
= 675 cm^3
= 675 x 10^-3 m^3
Thermal conductivity of unknown material,k = = ?
Solution:
In thermal conduction,
Rate of energy transfer, P = k A ( 1 - 2 ) / L
m c / t = k A ( 1 - 2 ) / L
V c / t = A ( 1 - 2 ) / L
Substite the above values in this expression to find thermal conductivity
coefficient '' of the unknown material.
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