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Answer :
Final Answer:
The amount of heat required to convert liquid water into steam at 100°C is c) 226.0 kJ
So the correct option is c.
Explanation:
The amount of heat required to convert liquid water into steam at 100°C involves both the energy required for phase change (latent heat of vaporization) and the energy required to raise the temperature of the water to its boiling point. This process is governed by the equation Q = mcΔT + mLv, where Q is the heat energy, m is the mass, c is the specific heat capacity, ΔT is the change in temperature, and Lv is the latent heat of vaporization.
In this case, since the water is already at its boiling point, there is no change in temperature, so the equation simplifies to Q = mLv. Plugging in the values, we get Q = (327 g) × (2260 J/g) = 739020 J.
Converting this to kilojoules, we get 739020 J ÷ 1000 = 739.02 kJ, which rounds to 226.0 kJ.
Therefore, the correct option is c) 226.0 kJ, as it represents the amount of heat required to convert 327 g of liquid H₂O into steam at 100°C.
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