If you add 700 kJ of heat to 700 g of water originally at 70.0°C, how much water is left in the container? The latent heat of vaporization of water is 22.6 × $10^5$ J/kg, and its specific heat capacity is 4186 J/kg ∙ K.

Question

Quizplus · Accepted Answer

First, calculate the energy needed to raise the temperature of water from 70°C to 100°C: $ Q = mc\Delta T = 0.7 \, \text{kg} \times 4186 \, \text{J/kg} \cdot \text{K} \times 30 \, \text{K} = 87906 \, \text{J} $. Subtract this from 700 kJ to find energy available for vaporization: $ 700,000 \, \text{J} - 87906 \, \text{J} = 612094 \, \text{J} $. The mass of water vaporized is $ \frac{612094 \, \text{J}}{22.6 \times 10^5 \, \text{J/kg}} \approx 0.271 \, \text{kg} $. Subtract from initial mass: $ 0.7 \, \text{kg} - 0.271 \, \text{kg} = 0.429 \, \text{kg} $ or 429 g.

If You Add 700 KJ of Heat to 700 G 10510^5105

If You Add 700 KJ of Heat to 700 G $10^5$