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Numericals · Q26

Q.Specific latent heat of vaporization of water is 2.26 × 10⁶ J/kg. Calculate the energy needed to change 5.0 g of water into steam at 100 ºC.

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Given latent heat of vaporization of water L=2.26×106 J/kgL=2.26\times10^6\text{ J/kg} and mass m=5.0 g=5.0×10−3 kgm=5.0\text{ g}=5.0\times10^{-3}\text{ kg}, use Q=mLQ=mL (Eq. 7.33): Q=5.0×10−3×2.26×106=11300 J=11.3×103 JQ = 5.0\times10^{-3}\times2.26\times10^6 = 11300\text{ J} = 11.3\times10^3\text{ J}. This is the energy needed to convert the water into steam ONLY (at the boiling point, with no temperature change), not to first heat it up to 100 °C from some lower starting temperature -- the question specifies the water is already at 100 °C. [!ANSWER] Q = 11.3 × 10³ J

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