Answer:
the energy absorbed is 4.477 x 10⁶ J
Explanation:
mass of the liquid, m = 13 kg
initial temperature of the liquid, t₁ = 18 ⁰C
final temperature of the liquid, t₂ = 100 ⁰C
specific heat capacity of water, c = 4,200 J/kg⁰C
The energy absorbed is calculated as;
H = mcΔt
H = mc(t₂ - t₁)
H = 13 x 4,200(100 - 18)
H = 4.477 x 10⁶ J
Therefore, the energy absorbed is 4.477 x 10⁶ J
Uh.. what's the question..?
Water evaporates at 100⁰C
So change in temperature = 100-20 = 80⁰C
Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg
Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg
So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ
So amount of heat require to evaporate water = 334.88 kJ
Answer:
the correct answer is the 60
Answer:
Condensation.
Explanation:
The boiling point of water is much higher than that of either nitrogen or oxygen gas . So when the mixture is condensed to a temperature lower than
100°C , water vapor will come out first in the form of water leaving other
elements of mixture in gaseous phase. In this way, water vapor will get separated from others.
