The boiling point of water is 100°C. So at 101°C, the water is steam. Compute the specific heat first from 101 to 100.
E = mCΔT, where c for steam is 1.996 kJ/kg·°C
E₁ = (0.65 kg)(1.996 kJ/kg·°C)(101 - 100°C) = 1.2974 kJ
Next, let's solve the latent heat when steam turns to liquid. The heat of vaporization of water is 2260 kJ/kg.
E₂ = mHvap = (0.65 kg)(2260 kJ/kg) = 1469 kJ
Lastly, let's solve the energy to bring down the temperature to 51°C. The specific heat of liquid water is 4.187 kJ/kg·°C.
E₃ = (0.65 kg)(4.187 kJ/kg·°C)(100 - 51°C) = 139.36 kJ
Thus,
<em>Total energy = 1.2974 kJ+1469 kJ+139.36 kJ = 1,609.66 kJ</em>
Equilibrium<span>-the condition of a system when no observable change is taken place or the kinetic energy is equal. </span>Equilibrium means<span> to stay balanced or equal.</span>
I think transfers is the answer
Answer:
0.023 Ohms
Explanation:
Given data
Length= 2.8m
radius= 1.03mm
current I= 1.35 A
voltage V= 0.032V
We know that from Ohm's law
V= IR
Now R= V/I
Substitute
R= 0.032/1.35
R= 0.023 Ohms
Hence the resistance is 0.023 Ohms
