Bobby, a 100 kg football player has a fever of 29 degrees C. What mass of water must evaporate from his body to cool him to 27 degrees C?
heat= Hv*mass
masskid*cwater*2=masswater*Heatvaporization solve for mass water.
To find the mass of water that must evaporate from Bobby's body to cool him to 27 degrees Celsius, we need to use the specific heat capacity formula:
Q = mcΔT
where:
- Q is the heat transferred
- m is the mass of the substance (water in this case)
- c is the specific heat capacity of the substance
- ΔT is the change in temperature
First, let's calculate the heat transferred (Q). Since the body is losing heat, we use a negative sign:
Q = -mcΔT
The specific heat capacity of water is approximately 4.18 J/g°C (joules per gram per degree Celsius). However, we need to convert the mass of Bobby from kilograms to grams:
Bobby's mass = 100 kg = 100,000 g
Next, let's calculate the change in temperature (ΔT):
ΔT = final temperature - initial temperature
ΔT = 27°C - 29°C
ΔT = -2°C
Now, we can substitute the values into the formula:
Q = -(100,000 g)(4.18 J/g°C)(-2°C)
Simplifying,
Q = 836,000 J
Since evaporation does not require a change in temperature, all the heat transferred goes into phase change. The heat of vaporization (latent heat) for water is approximately 2,260 J/g. Let's calculate the mass of water using this information:
Q = mL
m = Q / L
m = 836,000 J / 2,260 J/g
m ≈ 369.47 g
Therefore, approximately 369.47 grams of water must evaporate from Bobby's body to cool him from 29°C to 27°C.