Deci, Centi, then Nano is the correct order from largest to smallest
Answer:
No, it is not proper to use an infinitely long cylinder model when finding the temperatures near the bottom or top surfaces of a cylinder.
Explanation:
A cylinder is said to be infinitely long when is of a sufficient length. Also, when the diameter of the cylinder is relatively small compared to the length, it is called infinitely long cylinder.
Cylindrical rods can also be treated as infinitely long when dealing with heat transfers at locations far from the top or bottom surfaces. However, it not proper to treat the cylinder as being infinitely long when:
* When the diameter and length are comparable (i.e have the same measurement)
When finding the temperatures near the bottom or top of a cylinder, it is NOT PROPER TO USE AN INFINITELY LONG CYLINDER because heat transfer at those locations can be two-dimensional.
Therefore, the answer to the question is NO, since it is not proper to use an infinitely long cylinder when finding temperatures near the bottom or top of a cylinder.
Answer:
The greater the amplitude the greater the energy.
(Think of a water wave - which carries greater energy a 1 ft wave or
a 10 ft wave)
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
