Your hep please as I ask with respect .
a. What mass of water must evaporate from the skin of a 70kg man to cool his his body 1.00°C? The heat of vaporisation of water at body temperature (37°C) is 2.42×10^6J/Kg×K. The specific heat of at typical human body is 3480J/Kg×K
b. What volume of water must the man drink to replenish the evaporated water? Compare to the volume of the soft-drink can(355cm^3)?
a. Why did the man ask the water to cool his body? Because he didn't want to be called a "hot mess!" Let's calculate the amount of water that needs to evaporate.
To calculate the mass of water that must evaporate, we can use 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
ΔT is the change in temperature
We know that Q is equal to the heat of vaporization, so we can rearrange the formula to solve for mass:
m = Q / (c * ΔT)
Plugging in the values, we get:
m = (2.42×10^6 J/kg * K) / (3480 J/kg * K * 1 K)
Simplifying the equation will give us the mass of water that needs to evaporate.
b. Now, let's calculate the volume of water the man needs to drink to replenish the evaporated water. Shall we?
To calculate volume, we can use the formula:
V = m / ρ
Where:
V is the volume
m is the mass
ρ is the density of water
Since the density of water is approximately 1000 kg/m³, we can plug in the values and calculate the volume.
V = m / ρ
V = (mass of water evaporated) / 1000 kg/m³
Now, let's compare this volume to a soft drink can. Why did the man compare the volume to a soft-drink can? Because he wanted to make sure he wasn't drinking like a "fish out of a can!" So, let's see if the volume is similar to the can.
Is there anything else I can humorously assist you with?
a. To calculate the mass of water that must evaporate from the skin of the man to cool his body by 1.00°C, we can use the formula:
Heat transferred = mass of water * specific heat * temperature change
Given:
- Mass of the man (m) = 70 kg
- Specific heat of water (c) = 3480 J/kg·K
- Temperature change (ΔT) = 1.00 °C
Since the heat transferred is equal to the heat of vaporization of water (2.42 × 10^6 J/kg), we can equate the two equations:
Heat transferred = Heat of vaporization
mass of water * specific heat * temperature change = heat of vaporization
Rearranging the equation, we can solve for the mass of water:
mass of water = heat of vaporization / (specific heat * temperature change)
mass of water = (2.42 × 10^6 J/kg) / (3480 J/kg·K * 1.00 °C)
mass of water = 694.83 kg
Therefore, approximately 694.83 kg of water must evaporate from the skin of the man to cool his body by 1.00°C.
b. To calculate the volume of water the man needs to drink to replenish the evaporated water, we can assume that the density of water is approximately 1 kg/cm³.
Given:
- Mass of water evaporated (m) = mass of water calculated in part a = 694.83 kg
- Density of water (ρ) = 1 kg/cm³
We can use the formula:
Volume of water = mass of water / density of water
Volume of water = 694.83 kg / (1 kg/cm³)
Since we want the answer in cm³, we do not need to convert the units.
Volume of water = 694.83 cm³
Therefore, the man needs to drink approximately 694.83 cm³ (or mL) of water to replenish the evaporated water.
Comparing this to the volume of the soft drink can, which is 355 cm³, we can see that the volume of water the man needs to drink is significantly larger, more than twice the volume of the soft drink can.
a. To find the mass of water that must evaporate from the skin of the man to cool his body by 1.00°C, we can use the equation:
Q = m * ΔT * c
Where:
Q is the heat transferred (in joules)
m is the mass of water evaporated (in kilograms)
ΔT is the change in temperature (in kelvin)
c is the specific heat capacity of water (in joules per kilogram per kelvin)
In this case, the change in temperature is 1.00°C, which is equivalent to 1.00 Kelvin. The specific heat capacity of water is 3480 J/kg·K.
Now, let's calculate the heat transferred using the equation:
Q = m * ΔT * c
Q = m * 1.00 * 3480
We know that the heat of vaporization of water at body temperature is 2.42 × 10^6 J/kg·K. This means that to vaporize 1 kilogram of water at body temperature, we need 2.42 × 10^6 joules of energy.
Since the heat transferred (Q) must be equal to the heat of vaporization, we can set up the equation:
m * 1.00 * 3480 = 2.42 × 10^6
Now we can solve for m:
m = (2.42 × 10^6) / (1.00 * 3480)
m ≈ 694.83 kg
Therefore, approximately 694.83 kilograms of water must evaporate from the man's skin to cool his body by 1.00°C.
b. To find the volume of water the man must drink to replenish the evaporated water, we need to use the conversion factor between mass and volume for water. The density of water is approximately 1000 kg/m^3.
We have already determined that the mass of evaporated water is approximately 694.83 kg. Now we can calculate the volume using the equation:
V = m / ρ
Where:
V is the volume of water (in cubic meters)
m is the mass of water (in kilograms)
ρ is the density of water (in kilograms per cubic meter)
V = 694.83 / 1000
V ≈ 0.69483 m^3
To compare it to the volume of a soft drink can, we can convert cubic meters to cubic centimeters:
1 cubic meter = 1,000,000 cubic centimeters
V ≈ 0.69483 * 1,000,000
V ≈ 694,830 cm^3
Therefore, the volume of water the man must drink to replenish the evaporated water is approximately 694,830 cubic centimeters (or 694.83 liters). Comparing it to a standard soft drink can with a volume of 355 cm^3, we can see that the man must drink approximately 1961 cans of soft drink to replenish the evaporated water.