How much energy would your body need to provide for the following :
a)sweating and evaporation of 1kg of water from your skin.
b)heat loss by radiation at the rate of 100 wats for 12 hours.
(the specific latent heat of vaporisation of water is 2.3MJkg-1)
a) Multiply your heat of vaporization number by 1 kg.
That is the answer in Megajoules.
Multiply by 1 million for joules
b) 12 hours is 43,200 seconds.
Multiply that by 100 joules/s for the answer in joules
2.3 Million Joules is 550 kcal or 550 "food Calories"
Fortunately, we burn a lot of Calories doing nothing
To calculate the energy needed for each of these scenarios, we'll use the relevant equations and principles of thermodynamics.
a) Sweating and evaporation of 1 kg of water from your skin:
The energy required for the evaporation of water can be determined by calculating the latent heat of vaporization. The specific latent heat of vaporization of water is given as 2.3 MJ/kg.
The energy needed for the evaporation of 1 kg of water can be calculated by multiplying the mass (1 kg) by the specific latent heat of vaporization (2.3 MJ/kg):
Energy = 1 kg * 2.3 MJ/kg
Energy = 2.3 MJ
Therefore, the body would need 2.3 MJ (megajoules) of energy for sweating and evaporation of 1 kg of water from your skin.
b) Heat loss by radiation at the rate of 100 watts for 12 hours:
To determine the energy required for heat loss by radiation, we need to consider the rate of heat loss (100 watts) and the duration (12 hours).
Energy can be calculated by multiplying the power (100 W) by the time (12 hours), but first, we need to convert the time to seconds:
12 hours * 60 minutes/hour * 60 seconds/minute = 43200 seconds
Now, we can calculate the energy:
Energy = Power * Time
Energy = 100 W * 43200 s
Energy = 4320000 joules
Therefore, the body would need 4320000 joules of energy for heat loss by radiation at the rate of 100 watts for 12 hours.