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?
To calculate the mass of water that needs to evaporate from Bobby's body to cool him down, we need to use the specific heat formula:
q = m * c * ΔT
Where:
q = heat transferred
m = mass of the substance (water)
c = specific heat capacity of the substance (water)
ΔT = change in temperature
First, we need to calculate the heat transferred when cooling Bobby from 29°C to 27°C. To do this, we need the specific heat capacity of water, which is approximately 4.18 J/g°C.
ΔT = final temperature - initial temperature
= 27°C - 29°C
= -2°C
Now we can calculate the heat transferred using the formula:
q = m * c * ΔT
Since we want to know the mass of water that needs to evaporate, we can rearrange the formula as follows:
m = q / (c * ΔT)
Substituting the values:
q = m * c * ΔT
m = q / (c * ΔT)
m = q / (4.18 J/g°C * -2°C)
Note: I converted the mass from kg to grams to maintain consistent units.
To make this calculation, we need the heat transferred (q). However, the specific heat capacity of the human body varies, so we cannot directly calculate "q" without more information or making assumptions. The calculation also assumes that the heat required to evaporate the water comes solely from Bobby's body.
If you have additional information or assumptions, please provide them, and we can proceed with a more accurate calculation.