How much energy is required to vaporize 14.27 grams of water at its boiling point? *
1 answer:
Energy required to vaporize : 32.3 kJ
<h3>Further explanation </h3>The heat to change the phase can be formulated :
Q = mLf (melting/freezing)
Q = mLv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
Mass of water = 14.27 g
Latent heat of vaporization ( boiling point of 100 ºC) : 2260 J/g
Energy required :
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Answer:
Heat transfer during melting of ice plays greater role in cooling of liquid water.
Explanation:
Temperature of ice = -10 °c
Temperature of water = 0 °c
When ice cube is dipped in to the water.the heat transfer
Q = m c ΔT
⇒ Q = 1 × 2.01 × 10
⇒ Q = 20.1 KJ
Heat transfer during melting of ice = latent heat of ice
Latent heat of ice = 334 KJ
⇒ = 334 KJ
Heat transfer during melting of ice is greater value than heat transfer during warming of ice from -10°C to 0°C.
Thus heat transfer during melting of ice plays greater role in cooling of liquid water.