**Calculating Heat Required for Phase Change: Water to Steam**To determine the heat required to convert 10.0 g of water at 100°C to steam at 100.0 °C, we need to consider the energy needed for the phase change from liquid to gas.**Given Data:**- Mass of water = 10.0 g- Temperature = 100°C- Specific heat of water = 1.00 cal/(g × °C)- Heat of fusion (melting ice) = 80.0 cal/g (not required here)- Heat of vaporization = 540.0 cal/g**Calculation:**Only the heat of vaporization is needed as the water is already at 100°C, the boiling point. The energy required is calculated using the formula:Q = m × Hvwhere:- Q = heat energy (calories)- m = mass of the substance (g)- Hv = heat of vaporization (cal/g)**Substitute the values:**Q = 10.0 g × 540.0 cal/g = 5400.0 calTherefore, 5400.0 calories of energy are required to convert 10.0 g of water at 100°C to steam at 100°C.

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Calculate the heat required to convert 10.0 g of water at 100°C to steam at 100.0 °C. The specific heat of water is 1.00 cal/(g x °C); the heat of fusion is 80.0 cal/g; and the heat of vaporization is 540.0 cal/g.

Calculate the heat required to convert 10.0 g of water at 100°C to steam at 100.0 °C. The specific heat of water is 1.00 cal/(g x °C); the heat of fusion is 80.0 cal/g; and the heat of vaporization is 540.0 cal/g.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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