Answer:
A)
It should be Non- toxic
It should possess high Thermal conductivity
It should have the Required Thermal diffusivity
B)
- stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
- porcelain: mostly used for mugs and it is non-toxic
- Pyrex : posses good thermal conductivity used in oven
C) All the materials are suitable because they serve different purposes when making modern kitchen cookware
Explanation:
A) characteristics required of a ceramic material to be used as a kitchen cookware
- It should be Non- toxic
- It should possess high Thermal conductivity
- It should have the Required Thermal diffusivity
B) comparison of three ceramic materials as to their relative properties
- stoneware : This material has good thermal diffusivity and it is quite affordable and it is used in making pizza stones
- porcelain: mostly used for mugs and it is non-toxic
- Pyrex : posses good thermal conductivity used in ovens
C) material most suitable for the cookware.
All the materials are suitable because they serve different purposes when making modern kitchen cookware
A clean machine is a clean machine :-)
Explanation:
The wind is an actual form of solar energy. winds are caused by the heating of the atmosphere by the sun, the rotation of the earth, and the earth's surface irregularities. The wind is capture in a wind turbine which provides a renewable energy source, the wind makes the rotor spin, as the rotor spins the movement of the blades drives a generator that creates energy, also known as wind power. The average wind efficiency of turbines is between 35-45%.
Advantages of wind power
- Wind power is cost-effective
- wind creates jobs
- wind enables US industry growth and US competitiveness
-it's a clean fuel source
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
