(a) How many kilograms of water must evaporate from a 60.0 kg woman to lower her body temperature by 0.750°C?
For human body c=3470 J/kg.
At 37.0ºC the heat of vaporization Lv for water is 2430 kJ/kg or 580 kcal/kg
m•c•Δt= m(cond) •L(v)
m(cond) = m•c•Δt/L(v) =
=60•3470•0.75/2430000 =6.43•10⁻² kg=64.3 g
To determine the amount of water that needs to evaporate from a 60.0 kg woman to lower her body temperature by 0.750°C, we can use the specific heat capacity of water and the formula:
Q = m * c * ΔT
where:
Q is the heat absorbed or released
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we want to find the mass (m) of water that needs to evaporate, so we can rearrange the formula:
m = Q / (c * ΔT)
To calculate the value of Q, we need the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g°C.
Now we can substitute the values into the formula:
m = Q / (c * ΔT)
= Q / (4.18 J/g°C * 0.750°C)
To find the value of Q, we need to use another formula:
Q = mcΔT
The value of Q is the heat absorbed or released, which in this case is dependent on the latent heat of vaporization, L, of water. The latent heat of vaporization, L, is the amount of heat energy required to change a given mass (m) of a substance from its liquid phase to its gaseous phase without changing its temperature. For water, the value of L is approximately 2.26 x 10^6 J/kg.
Now, we can plug in the values into the formula:
Q = mcΔT
Q = (60.0 kg) * (4.18 J/g°C) * (0.750°C)
Finally, we can substitute the value of Q into the first formula to find the mass of water:
m = Q / (c * ΔT)
= [(60.0 kg) * (4.18 J/g°C) * (0.750°C)] / (4.18 J/g°C * 0.750°C)
Simplifying the equation:
m = 60.0 kg
Therefore, to lower the body temperature of a 60.0 kg woman by 0.750°C, no water needs to evaporate.
To calculate the amount of water that needs to evaporate, we need to use the equation:
Heat absorbed = mass × specific heat × temperature change
First, we need to find the heat absorbed by the woman's body. The equation for heat absorbed is:
Heat absorbed = mass × specific heat × temperature change
The specific heat of water is 4.18 J/g°C, and we are looking for the mass of water in kilograms that needs to evaporate.
Rearranging the equation:
mass = Heat absorbed / (specific heat × temperature change)
To find the heat absorbed, we use the equation:
Heat absorbed = mass × latent heat of vaporization
The latent heat of vaporization represents the amount of heat required to change one kilogram of a substance from liquid to gas at its boiling point. For water, the latent heat of vaporization is approximately 2.26 × 10^6 J/kg.
Now we can calculate the heat absorbed:
Heat absorbed = (60.0 kg) × (2.26 × 10^6 J/kg) = 1.356 × 10^8 J
Now we can substitute this value back into the equation for mass:
mass = (1.356 × 10^8 J) / (4.18 J/g°C × 0.750°C)
mass ≈ 6.79 × 10^4 g
Converting grams to kilograms:
mass ≈ 67.9 kg
Therefore, approximately 67.9 kilograms of water must evaporate from a 60.0 kg woman to lower her body temperature by 0.750°C.