Question

How much heat (in kj) is required to convert 429 g of liquid H₂O at 21.7°c into steam at 144°c? (assume that the specific heat of liquid water is 4.184 j/g·°c, the specific heat of steam is 2.078 j/g·°c, and that both values are constant over the given temperature ranges. the normal boiling point of H₂O is 100.0°c. the enthalpy of vaporization for water at 100.0°c is δhvap = 40.657 kj/mol.)

Answer 1

To convert 429 g of liquid H₂O at 21.7 °C into steam at 144 °C, we need to consider three steps: heating the liquid water, boiling the liquid water into steam, and heating the steam to the final temperature.

Step-by-step explanation:

To calculate the heat required to convert liquid H₂O at 21.7 °C into steam at 144 °C, we need to consider three steps: heating the liquid water to its boiling point, boiling the liquid water into steam, and heating the steam to the final temperature.

The first step is heating the liquid water. We can use the equation Q = m * C * ∆T, where Q is the heat, m is the mass, C is the specific heat, and ∆T is the change in temperature. The heat required for this step is (429 g) * (4.184 J/g·°C) * (100.0 °C - 21.7 °C).

The second step is boiling the liquid water into steam. We need to multiply the mass of liquid water by the enthalpy of vaporization, which is given as 40.657 kJ/mol. We can convert grams to moles using the molar mass of water, and then multiply by the enthalpy of vaporization. The heat required for this step is (429 g) / (18.015 g/mol) * (40.657 kJ/mol).

The third step is heating the steam to the final temperature. We can use the same equation as in the first step, using the specific heat of steam, which is given as 2.078 J/g·°C. The heat required for this step is (429 g) * (2.078 J/g·°C) * (144 °C - 100 °C).

To calculate the total heat required, we sum up the heat required for each step.