During a strenuous workout, a student generates 2000 kJ of heat energy. What mass of water would have to evaporate from the student's skin to dissipate this much heat?

2,000,000 = mass water x delta Hvap.

Look up delta Hvap and solve for mass water, in grams. Be sure delta Hvap is in joules or convert to joules.

Well, let me put it this way. If the student wants to become a human sauna, they'd have to evaporate a lot of water! But let's crunch the numbers for some giggles. To figure out the mass of water that would need to evaporate, we can use the equation: Q = m * ΔHv, where Q is the heat energy, m is the mass, and ΔHv is the heat of vaporization of water (2.26 kJ/g). Rearranging the equation, we get m = Q / ΔHv. Substituting the values, we have m = 2000 kJ / 2.26 kJ/g ≈ 884.96 g. So, to dissipate that much heat, the student would have to evaporate approximately 885 grams of sweat. That's a lot of waterworks!

To determine the mass of water that would have to evaporate from the student's skin, we can use the equation for specific heat:

Q = m * c * ΔT

where:
Q = heat energy in joules
m = mass of water in kilograms
c = specific heat capacity of water (4.18 kJ/kg°C)
ΔT = change in temperature

First, let's convert the heat energy from kilojoules to joules:
2000 kJ * 1000 = 2,000,000 J

Next, we need to determine the change in temperature. Assuming the student's body temperature is 37°C and the room temperature is 25°C, the change in temperature would be:
ΔT = 37°C - 25°C = 12°C

Now, we can rearrange the equation to solve for the mass of water:
m = Q / (c * ΔT)

Plugging in the values:
m = 2,000,000 J / (4.18 kJ/kg°C * 12°C)
m = 119,617.22 kg (rounded to two decimal places)

Therefore, approximately 119,617.22 kilograms of water would need to evaporate from the student's skin to dissipate 2000 kJ of heat energy.

To determine the mass of water that would need to evaporate to dissipate 2000 kJ of heat energy, we need to use the concept of specific heat and heat of vaporization.

First, let's find the energy required to raise the temperature of the water on the student's skin from its initial temperature to its boiling point. The specific heat capacity of water is approximately 4.18 kJ/kg°C.

Let's assume the initial temperature of the water is 37°C (average body temperature) and its boiling point is 100°C.

The energy required to raise the temperature of a given mass of water can be calculated using the formula:

Energy = mass × specific heat capacity × change in temperature

Since the initial temperature of the water is 37°C, the change in temperature will be 100°C - 37°C = 63°C.

Now we can rearrange the formula to find the mass of water:

mass = Energy / (specific heat capacity × change in temperature)

mass = 2000 kJ / (4.18 kJ/kg°C × 63°C)
mass = 9.51 kg

Therefore, approximately 9.51 kg of water would need to evaporate from the student's skin to dissipate 2000 kJ of heat energy.