What are practical considerations you might encounter when you increase the moment of inertia (I) while keeping the cross-sectio
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Answer:
(a)
<em>d</em>Q = m<em>d</em>q
<em>d</em>q = <em>d</em>T
=
(T₂ - T₁)
From the above equations, the underlying assumption is that remains constant with change in temperature.
(b)
Given;
V = 2L
T₁ = 300 K
Q₁ = 16.73 KJ , Q₂ = 6.14 KJ
ΔT = 3.10 K , ΔT₂ = 3.10 K for calorimeter
Let be heat constant of calorimeter
Q₂ = ΔT
Heat absorbed by n-C₆H₁₄ = Q₁ - Q₂
Q₁ - Q₂ = m ΔT
number of moles of n-C₆H₁₄, n = m/M
ρ = 650 kg/m³ at 300 K
M = 86.178 g/mol
m = ρv = 650 (2x10⁻³) = 1.3 kg
n = m/M => 1.3 / 0.086178 = 15.085 moles
Q₁ - Q₂ = m ' ΔT
= (16.73 - 6.14) / (15.085 x 3.10)
= 0.22646 KJ mol⁻¹ k⁻¹
Answer with Explanation:
The capillary rise in 2 parallel plates immersed in a liquid is given by the formula
where
is the surface tension of the liquid
is the contact angle of the liquid
is density of liquid
'g' is acceleratioj due to gravity
'd' is seperation between thje plates
Part a) When the liquid is water:
For water and glass we have
Applying the values we get
Part b) When the liquid is mercury:
For mercury and glass we have
Applying the values we get
The negative sign indicates that there is depression in mercury in the tube.