Answer:
6.36 105 J
Explanation:
In calorimetry all the energy given to a system is converted to heat and the equation for heat is
Q = m ΔT
The temperature can be in degrees Celsius or Kelvin since the interval between them is the same, substitute and calculate
Q = 1.9 4186 (373-293)
Q = 6.36 105 J
This heat equals the energy supplied
This question is incomplete.Here is complete one
Consider a telescope with a small circular aperture of diameter 2.0 centimeters.
If two point sources of light are being imaged by this telescope, what is the maximum wavelength (lambda) at which the two can be resolved if their angular separation is 3.0x10^-5 radians? Express your answer in nanometers
Answer:
λ =492nm
Explanation:
Given Data
diameter=2.0 cm
angular separation=3.0x10^-5 radians
λ=?
Solution
sin(theta) = 1.22 x λ /D
λ=(0.02×sin(3.0×10⁻⁵))/1.22
λ=492nm
Answer:
The correct answer is option B) Mechanical waves tend to have longer wavelengths.
Explanation:
Before we begin, let's clarify what these two types of waves are.
Both electromagnetic waves and mechanical waves are classified by their ability to transmit energy in empty space.
<u>Electromagnetic waves are capable of doing that, but mechanical waves are not.</u>
The light waves that reach our planet are an example of electromagnetic waves.
On the other hand, mechanical waves need a medium to be able to move energy, it is not the same as with electromagnetic waves that are capable of transporting it on their own.
An example of a mechanical wave is sound. Sound is not capable of traveling in empty space.
This leads us to the conclusion that Mechanical waves tend to have longer wavelengths.
True and you need to do your reasearch more.
Answer:
The required mass of ice is 12.5 kg.
Explanation:
Mass of hot tea, = 1.8 kg
Initial temperature of hot tea = 80°C = 353 K
Initial temperature of ice = 0.00°C = 273 K
Final temperature of mixture = 10°C = 283 K
Heat of fusion of ice, L = 334 kJ/kg
Specific heat capacity of tea, = Specific heat capacity of water = 4190 J/kg/K
Heat lost tea = Heat gained by ice
Δ = L + Δ
Δ = (L + Δ)
1.8 x 4190 x (353 - 283) = (334 + (4190 x (283 - 2730))
1.8 x 4190 x 70 = (334 + (4190 x 10))
527940 = 42234
=
= 12.5004
= 12.5 kg
The required mass of ice is 12.5 kg.