During a strenous workout, an athlete generates 1890.0 kj of heat energy. What mass of water would have to evaporate from the athlete's skin to dissipate this much heat?

To find the mass of water that would have to evaporate from the athlete's skin to dissipate the heat energy, we can use the equation:

Q = m * ΔHv

Where:
Q = heat energy (in joules)
m = mass of water (in kilograms)
ΔHv = heat of vaporization of water (in joules per kilogram)

Since the heat energy is given in kilojoules, we need to convert it to joules:

1 kilojoule = 1000 joules

So, 1890.0 kJ = 1890.0 * 1000 J = 1,890,000 J

Now, we need to determine the heat of vaporization of water. The heat of vaporization of water is approximately 2260 kJ/kg.

1 kilojoule = 1000 joules

So, 2260 kJ/kg = 2260 * 1000 J/kg = 2,260,000 J/kg

Now we can substitute the values into the equation and solve for the mass (m):

1,890,000 J = m * 2,260,000 J/kg

To isolate m, we divide both sides of the equation by 2,260,000 J/kg:

m = 1,890,000 J / 2,260,000 J/kg

m ≈ 0.836 kg

Therefore, approximately 0.836 kilograms of water would have to evaporate from the athlete's skin to dissipate this much heat.

To determine the mass of water that would need to evaporate from the athlete's skin to dissipate 1890.0 kJ of heat energy, we can use the principle of specific heat capacity and the heat of vaporization of water.

1. Calculate the heat required to evaporate the water:
- The heat of vaporization of water is approximately 2260 kJ/kg (varies slightly with temperature).
- So, using this value, divide the total heat energy (1890.0 kJ) by the heat of vaporization to find the mass of water required:
Mass of water = Heat energy / Heat of vaporization

2. Substitute the values into the equation:
Mass of water = 1890.0 kJ / 2260 kJ/kg

3. Calculate the mass of water:
Divide the heat energy (in kJ) by the heat of vaporization (in kJ/kg) to get the mass in kg.

Remember to use consistent units throughout the calculations for accurate results.

Let's calculate the mass of water required to dissipate 1890.0 kJ of heat energy using the formula above.