What is the total heat flow if 27 grams of steam at 143°C is cooled to become water at 34°C? The specific heat of water is 4.18 J/g • °C; the specific heat of steam is 2.0 J/g • °C.The heat of vaporization of water is 2.26 kJ/g,and the boiling point of water is 100°C.

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The total heat flow includes cooling the steam to 100°C, condensing the steam to water, and then cooling the water to 34°C. Calculate each step and sum them:  1. Cooling steam: $ q_1 = m \cdot c_{\text{steam}} \cdot \Delta T = 27 \, \text{g} \cdot 2.0 \, \text{J/g°C} \cdot (143°C - 100°C) = 2322 \, \text{J} $  2. Condensing steam: $ q_2 = m \cdot \text{heat of vaporization} = 27 \, \text{g} \cdot 2260 \, \text{J/g} = 61020 \, \text{J} $  3. Cooling water: $ q_3 = m \cdot c_{\text{water}} \cdot \Delta T = 27 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot (100°C - 34°C) = 7471.56 \, \text{J} $  Total heat flow: $ q_{\text{total}} = q_1 + q_2 + q_3 = 2322 + 61020 + 7471.56 = 70813.56 \, \text{J} \approx 71 \, \text{kJ} $

What Is the Total Heat Flow If 27 Grams of Steam