Final answer:
To heat 10 grams of water from 50 ºC to 100 ºC and then vaporize it requires 24.7 kJ of heat. This includes heating the liquid water up to boiling point and then supplying the heat of vaporization to convert the liquid to vapor.
Step-by-step explanation:
To calculate how much heat is required to heat 10 grams of liquid water from 50 ºC to vapor at 100 ºC, we need to consider two processes: heating the liquid water from 50 ºC to 100 ºC and then vaporizing it at 100 ºC. The specific heat capacity of water is 4.184 J/g°C, which tells us how much heat is needed to raise the temperature of 1 gram of water by 1°C, and the heat of vaporization of water at 100 ºC is approximately 2260 J/g, which is the heat required to turn 1 gram of liquid water into vapor at its boiling point.
First, we calculate the heat required to raise the temperature of 10 grams of water from 50 ºC to 100 ºC using the formula q = m × c × ΔT, where q is the heat absorbed, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. For this step, q = 10 g × 4.184 J/g°C × (100 ºC - 50 ºC) = 2092 J.
Next, we calculate the heat required to vaporize the water at 100 ºC using the formula q = m × ΔHvap, where ΔHvap is the heat of vaporization. For this step, q = 10 g × 2260 J/g = 22600 J.
Finally, we add both quantities of heat and convert the sum into kilojoules (kJ) by dividing by 1000: (2092 J + 22600 J) / 1000 = 24.692 kJ, which we can round to one decimal place as 24.7 kJ. Therefore, the heat required is 24.7 kJ.