Physics · Heat, temperature, thermal expansion, specific heat capacity, calorimetry, change of state, latent heat
The heat energy required to vaporize 5 kg of water at is nearly (Latent heat of
The heat energy required to vaporize 5 kg of water at \( 373 \mathrm{K} \) is nearly (Latent heat of vaporization \( \left.\boldsymbol{L}_{\boldsymbol{v}}=\mathbf{2 2 7 0} \boldsymbol{J}\right) \)
- A. 2700 K.cal
- B. 1000 K.cal
- C. 27 к.са
- D. 270 K.cal
Step-by-step solution
The heat required is Q = m * L_v = 5 kg * 2270 J/g (assuming L_v = 2270 J/g, which is equivalent to 2270 kJ/kg since 1 kg = 1000 g, but careful: 2270 J/g = 2.27e6 J/kg). So Q = 5000 g * 2270 J/g = 1.135e7 J. Convert to kilocalories: 1 kcal = 4184 J, so Q = 1.135e7 / 4184 ≈ 2712 kcal ≈ 2700 kcal.
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