Suppose 4 g of water at 100°C is spread over a large surface so that 1 g evaporates rapidly. If evaporation of the 1 gram takes 2,259 joules from the remaining 3 g of water and no other heat transfer takes place, what are the temperature and phase of the remaining 3 g once the 1 gram is completely evaporated?

Question

Quizplus · Accepted Answer

First, we can use the formula Q = mL to find the amount of heat required to evaporate 1 gram of water:
Q = (1 g) (2260 J/g) = 2260 J

Since the heat transfer takes place only between the 1 gram evaporated and the remaining 3 grams of water, we can set up an equation using the specific heat capacity of water:

Q = mcΔT
2260 J = (3 g)(4.184 J/g°C)(ΔT)
ΔT = 170.9°C

Subtracting this from the original temperature of 100°C, we get:
100°C - 170.9°C = -70.9°C

This means that the remaining 3 grams of water will be at -70.9°C, which is well below the freezing point of water. Therefore, the water will be in the solid phase, or ice, at this temperature. Thus, the correct choice is D, "The remaining 3 g will turn to 0°C ice."

Suppose 4 G of Water at 100°C Is Spread Over