Tony£¬a 110kg football player£¬has a fever of 39celcius. What mass of water must evaporate from his body to cool him to 37celcius£¿
assuming he is mostly water,
heat= 110*cwater*2
but heat also = masswaterevap*Lv
set them equal, and solve.
To determine the mass of water that needs to evaporate from Tony's body in order to cool him to 37 degrees Celsius, we can use the specific heat capacity equation:
Q = mcΔT
where:
Q = heat transferred (in joules)
m = mass of the substance (in this case, water)
c = specific heat capacity of water (approximately 4.18 J/g°C)
ΔT = change in temperature (in this case, the difference between 39°C and 37°C)
First, let's calculate the heat transferred:
Q = mcΔT
Since the heat transferred is equal to the heat absorbed by Tony's body, we can assume that:
Q = mwater × cwater × ΔTwater
Now we need to convert the mass of Tony (given in kilograms) to grams:
mplayer = 110 kg = 110,000 g
We also need to calculate the temperature difference:
ΔTwater = 39°C - 37°C = 2°C
Now we can rearrange the equation to solve for the mass of water:
mwater = Q / (cwater × ΔTwater)
Calculating the heat transferred (Q):
Q = mplayer × cplayer × ΔTplayer
Since we know the specific heat capacity of the human body is approximately 3.5 J/g°C, we have:
Q = 110,000 g × 3.5 J/g°C × (39°C - 37°C)
Q = 110,000 g × 3.5 J/g°C × 2°C
Q = 770,000 J
Now we can substitute the calculated values into the equation:
mwater = 770,000 J / (4.18 J/g°C × 2°C)
mwater ≈ 92,344.5 g
Therefore, approximately 92,344.5 grams (or 92.34 kg) of water must evaporate from Tony's body to cool him from 39°C to 37°C.