Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x ]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
Answer: -112200J
Explanation:
The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of water vapour = 30.0g
C = 187 J/ G°C
Φ = (Final temperature - Initial temperature)
= 100°C - 120°C = -20°C
Then apply the formula, Q = MCΦ
Q = 30.0g x 187 J/ G°C x -20°C
Q = -112200J (The negative sign does indicates that heat was released to the surroundings)
Thus, -112200 joules of heat is released when cooling the superheated vapour.
The magnetic fields are the reason why magnets repel and attract. eachother.
