How much heat is required to vaporize 50.0 g of water if the initial temperature of the water is 25.0 $\circ$ C and the water is heated to its boiling point, where it is converted to steam? The specific heat capacity of water is 4.18 J.( $\circ$ C)^-¹g^-¹ and the standard enthalpy of vaporization of water at its boiling point is 40.7 kJ.mol^-¹.

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Quizplus · Accepted Answer

To find the total heat required, we need to calculate both the heat needed to raise the temperature of water from 25°C to 100°C and the heat required to vaporize the water at 100°C. First, calculate the heat to raise the temperature: $q = mc\Delta T = 50.0\,g \times 4.18\,J/g°C \times (100°C - 25°C) = 50.0\,g \times 4.18\,J/g°C \times 75°C = 15,675\,J = 15.675\,kJ$. Then, calculate the heat to vaporize the water: $50.0\,g \times \frac{1\,mol}{18.015\,g} \times 40.7\,kJ/mol = 2.775\,mol \times 40.7\,kJ/mol = 112.98\,kJ$. Adding both gives $15.675\,kJ + 112.98\,kJ = 128.655\,kJ$, which is not an option due to a mistake in calculation or interpretation. The correct approach involves calculating the heat to raise the temperature to 100°C and then the heat for the phase change, but the provided options and calculations mismatch. The correct calculation for vaporization alone, considering 50.0 g of water, should directly involve the enthalpy of vaporization and the amount of water in moles, indicating a potential misunderstanding in applying these values directly to the options provided.

How Much Heat Is Required to Vaporize 50 ∘\circ∘ C and the Water Is Heated to Its Boiling Point

How Much Heat Is Required to Vaporize 50 $\circ$ C and the Water Is Heated to Its Boiling Point