Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, m₁ = 581 g
The second (mass) balance reading (of the water) in grams, m₂ = 526 g
The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ
The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ΔQ/Δm
∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g
Periods<span> going left to right. The periodic table also has a special name </span>for<span> its vertical columns. Each column </span>is<span> called a </span>group. The elements in eachgroup<span> have the same number </span>of<span> electrons </span>in the<span> outer orbital.</span>
Answer:
Explanation:
charge on the capacitor = capacitance x potential
= 1.588 x 3.4
= 5.4 C
Energy of capacitor = 1 / 2 C V ² , C is capacitance , V is potential
= .5 x 3.4 x 1.588²
= 4.29 J
If I be maximum current
energy of inductor = 1/2 L I² , L is inductance of inductor .
energy of inductance = Energy of capacitor
1/2 L I² = 4.29
I² = 107.25
I = 10.35 A
Time period of oscillation
T = 2π √ LC
=2π √ .08 X 3.4
= 3.275 s
current in the inductor will be maximum in T / 4 time
= 3.275 / 4
= .819 s.
Total energy of the system
= initial energy of the capacitor
= 4.29 J
