To solve this problem, we should recall the law of
conservation of energy. That is, the heat lost by the aluminium must be equal
to the heat gained by the cold water. This is expressed in change in enthalpies
therefore:
- ΔH aluminium = ΔH water
where ΔH = m Cp (T2 – T1)
The negative sign simply means heat is lost. Therefore we
calculate for the mass of water (m):
- 0.5 (900) (20 – 200) = m (4186) (20 – 0)
m = 0.9675 kg
Using same mass of water and initial temperature, the final
temperature T of a 1.0 kg aluminium block is:
- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)
- 900 T + 180,000 = 4050 T
4950 T = 180,000
T = 36.36°C
The final temperature of the water and block is 36.36°C
Answer:
λ = 102.78 nm
This radiation is in the UV range,
Explanation:
Bohr's atomic model for the hydrogen atom states that the energy is
E = - 13.606 / n²
where 13.606 eV is the ground state energy and n is an integer
an atom transition is the jump of an electron from an initial state to a final state of lesser emergy
ΔE = 13.606 (1 /
- 1 / n_{i}^{2})
the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon
DE = 13.606 (1/1 - 1/3²)
DE = 12.094 eV
let's reduce the energy to the SI system
DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J
let's find the wavelength is this energy, let's use Planck's equation to find the frequency
E = h f
f = E / h
f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴
f = 2.9186 10¹⁵ Hz
now we can look up the wavelength
c = λ f
λ = c / f
λ = 3 10⁸ / 2.9186 10¹⁵
λ = 1.0278 10⁻⁷ m
let's reduce to nm
λ = 102.78 nm
This radiation is in the UV range, which occurs for wavelengths less than 400 nm.
Explanation:
Always behave responsibly in the laboratory. Do not run around or play practical jokes. Always check the safety data of any chemicals you are going to use. Never smell, taste or touch chemicals unless instructed to do so.