Answer:
The upper and lower limits for the room-temperature thermal conductivity of a magnesium oxide material having a volume fraction of 0.10 of pores that are filled with still air are
Ku = 38.252 W/mK
K lower = 0.199 W/mK
Explanation:
As we know
Ku = Vp * Kair + Vmagnesium * K metal
Ku = 0.10 *0.02 + (1-0.25) * 51
Ku = 38.252 W/mK
The lower limit
K lower = Kmetal* Kair/( Vp * Kmetal + Vmetal * K air)
K lower = (0.02*51)/(0.10*51 + 0.90 * 0.02)
K lower = 0.199 W/mK
The initial volume is 116.65 mL
<u>Explanation:</u>
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Given:
Temperature, T₁ = 22°C
T₂ = 86°C
Volume, V₂ = 456 m
V₁ = ?
According to Charle's law:
Substituting the values:
Therefore, the initial volume is 116.65 mL
Answer: 150 grams
Explanation: m = V × ρ
= 15 milliliter × 10 gram/cubic centimeter
= 15 cubic centimeter × 10 gram/cubic centimeter
= 150 gram
