Question

Is it true that it takes more energy to vaporize \(1 \mathrm{kg}\) of saturated liquid water at \(100^{\circ} \mathrm{C}\) than it would at \(120^{\circ} \mathrm{C} ?\)

Short answer

Answer: No, it takes less energy to vaporize 1 kg of saturated liquid water at 120°C than at 100°C based on the specific latent heat of vaporization values and the assumption that it decreases with temperature following the Clausius-Clapeyron equation.

Step-by-step solution

Understand the concept of specific latent heat of vaporization

Specific latent heat of vaporization is the amount of energy required to change the phase of 1 kg of a substance from liquid to gas at constant temperature. This is an important concept in thermodynamics and will be used to assess the energy required for vaporizing water at different temperatures.

Collect the specific latent heat of vaporization values for water

The specific latent heat of vaporization for water can be found using a reference table or by consulting reliable sources. At 100°C, the specific latent heat of vaporization (L) for water is 2257 kJ/kg. Values for higher temperatures may not be readily available. If we want to find the value for 120°C, we can assume that the specific latent heat of vaporization decreases as temperature increases. In order to determine this, we can make use of the Clausius-Clapeyron equation to estimate the change in specific latent heat with temperature.

Using Clausius-Clapeyron equation to find specific latent heat at 120°C

The Clausius-Clapeyron equation relates the latent heat of vaporization, temperature, and vapor pressure of a substance. The equation is as follows: $$ \frac{d(ln(P))}{dT} = \frac{-L}{RT^2} $$ Where: - P is the vapor pressure - L is the specific latent heat of vaporization - R is the ideal gas constant (8.314 J/mol·K) - T is the absolute temperature in Kelvin To get the values of vapor pressure (P) at 100°C and 120°C, we can consult a steam table or a reliable source. We have: - P at 100°C = 101325 Pa - P at 120°C = 198390 Pa (Note that these values are approximated based on the saturated steam conditions) Now convert Celsius to Kelvin: - T1 = 100°C + 273.15 = 373.15 K - T2 = 120°C + 273.15 = 393.15 K With these values, we can calculate the specific latent heat of vaporization (L) at 120°C by integrating the Clausius-Clapeyron equation: $$ L = -RT^2 \int_{373.15 K}^{393.15 K} \frac{d(ln(P))}{dT} dT $$

Compare specific latent heat values at 100°C and 120°C

After determining the specific latent heat of vaporization for water at both 100°C and 120°C, we can compare the values to find out if it takes more energy to vaporize 1 kg of saturated liquid water at 100°C than at 120°C. Assuming that the specific latent heat of vaporization decreases with temperature following the Clausius-Clapeyron equation, we would find a smaller L at 120°C compared to 100°C.

Conclusion

Based on the specific latent heat of vaporization values for water at 100°C and 120°C, we can conclude that it takes less energy to vaporize 1 kg of saturated liquid water at 120°C than it would at 100°C.