Question

If 3.85 kJ of heat are added to water at 100.0°C, how much water is there?

Option 1: To raise the temperature of water at 100.0°C by 3.85 kJ, you'd need 1 gram of water.
Option 2: 3.85 kJ of heat can raise the temperature of 1 liter of water at 100.0°C.
Option 3: To heat 3.85 kg of water at 100.0°C, you'd require 1 kilojoule of energy.
Option 4: Adding 3.85 kJ of heat to water at 100.0°C would require 3.85 milliliters of water.

Answer 1

The correct calculation to determine the amount of water that can be vaporized by adding 3.85 kJ of heat is to divide the energy by the enthalpy of vaporization after converting kJ to J, resulting in the vaporization of approximately 1.7 grams of water.

Step-by-step explanation:

To determine how much water is involved when adding 3.85 kJ of heat to water at 100.0°C, we must use the concept of specific heat. The specific heat of water is 4.184 J/g °C, which means to raise the temperature of 1 gram of water by 1°C, we would need 4.184 J of energy. If the water is at 100.0°C, any additional heat added will not increase the water's temperature further but rather convert it to steam. The heat required for the phase change from liquid to gas is known as the enthalpy of vaporization, which is approximately 2260 kJ/kg for water.

None of the presented options correctly describes the relationship between the heat added and the amount of water. To calculate the amount of water that can be completely vaporized by 3.85 kJ of energy, we need to divide the energy amount by the enthalpy of vaporization (we would have to convert the energy units from kJ to J by multiplying by 1000 to match the units).

Thus, the correct calculation to find out how much water (in kilograms) can be vaporized by 3.85 kJ of heat would be:

• 3.85 kJ = 3850 J
• Amount of water vaporized = 3850 J / (2260 kJ/kg * 1000 J/kJ)
• Amount of water vaporized = 3850 J / 2260000 J/kg
• Amount of water vaporized = 0.00170354 kg, or about 1.7 grams of water.