Question

Our body's primary defense against high temperatures is through sweating (perspiration). Consider a 185 lb (84 kg) male farmer in the Central Valley of California on a July day when the air temperature is 110°F (43°C). On this day, the humidity is low enough such that all the sweat evaporates from the farmer's skin, enabling him to maintain a body temperature of 100°F (38°C). The farmer sweats 1 liter (1000 g of water) per hour, and 60% of the latent heat needed to evaporate this sweat is provided by the farmer's body (the other 40% is provided by the air). If the farmer suddenly dehydrates and stops sweating, how long will it take for his body temperature to rise to a near-fatal 110°F (43°C)? Please show all of your work. Assumptions: 1. It takes 4200 Joules of energy to warm each kilogram (kg) of his body 1°C (1.8°F). 2. The latent heat of vaporization is 2,400,000 J/kg. This value means that it takes 2,400,000 Joules of energy to convert one kilogram of sweat (a liquid) into a gas.

          Our body's primary defense against high temperatures is through sweating (perspiration). Consider a 185 lb (84 kg) male farmer in the Central Valley of California on a July day when the air temperature is 110°F (43°C). On this day, the humidity is low enough such that all the sweat evaporates from the farmer's skin, enabling him to maintain a body temperature of 100°F (38°C). The farmer sweats 1 liter (1000 g of water) per hour, and 60% of the latent heat needed to evaporate this sweat is provided by the farmer's body (the other 40% is provided by the air). If the farmer suddenly dehydrates and stops sweating, how long will it take for his body temperature to rise to a near-fatal 110°F (43°C)? Please show all of your work.

Assumptions: 
1. It takes 4200 Joules of energy to warm each kilogram (kg) of his body 1°C (1.8°F).
2. The latent heat of vaporization is 2,400,000 J/kg. This value means that it takes 2,400,000 Joules of energy to convert one kilogram of sweat (a liquid) into a gas.

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Added by Rachel B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Neelesh Sharma

03/28/2022

Step 1

Given: - Mass of the farmer = 84 kg - Specific heat capacity of the body = 4200 J/kg°C - Temperature difference = 43°C - 38°C = 5°C Total energy = mass x specific heat capacity x temperature difference Total energy = 84 kg x 4200 J/kg°C x 5°C Total energy = Show more…

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Our body's primary defense against high temperatures is through sweating (perspiration). Consider a 185 lb (84 kg) male farmer in the Central Valley of California on a July day when the air temperature is 110°F (43°C). On this day, the humidity is low enough such that all the sweat evaporates from the farmer's skin, enabling him to maintain a body temperature of 100°F (38°C). The farmer sweats 1 liter (1000 g of water) per hour, and 60% of the latent heat needed to evaporate this sweat is provided by the farmer's body (the other 40% is provided by the air). If the farmer suddenly dehydrates and stops sweating, how long will it take for his body temperature to rise to a near-fatal 110°F (43°C)? Please show all of your work. Assumptions: 1. It takes 4200 Joules of energy to warm each kilogram (kg) of his body 1°C (1.8°F). 2. The latent heat of vaporization is 2,400,000 J/kg. This value means that it takes 2,400,000 Joules of energy to convert one kilogram of sweat (a liquid) into a gas.

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Transcript

00:01 Hello friends, so we'll start with this is a thermal feed transfer position.

00:07 In this case it is given that a male having a mass of 84 kg on a july day, the air temperature is given as 1 .43 degrees celsius.

00:24 On this stage the humidity is low enough, we have to maintain body temperature at 38 degrees celsius.

00:33 The pharma sweats, 1 litre of water.

00:40 This is 1 liter of water per hour.

00:51 60 % of the latent heat needed to evaporate this sweat.

00:56 The other 40 % is provided by the air.

01:00 So we have to divide this part by part and is asking that the pharma suddenly dehydrates instead of treating.

01:09 So how long will it take his body to? will rise the temperature to 43 degrees celsius.

01:17 So if the body is not sweating, this temperature will rise up to 43 degrees celsius, that is the temperature of the air.

01:26 So we have to find this amount of heat.

01:30 First, that is expected because of this 100, 1 liter of water per hour.

01:36 This will extract some amount of heat.

01:39 Now, in the second part, when this is stopped, what will be the amount of heat? of temperature rise how much time this will take for the temperature to rise up to this position so we have to find the rate of speed transfer now amount of heat required to warm each kg of weight by 1 degree celsius is 4200 feet this is given already now the amount of late heat used to evaporate body is 60 % so 60 % is the latent heat 40 % is the sensibly, 43 minus 38, that is the 5 degree temperature difference maintained by this.

02:29 So the energy required to raise the temperature for 1 kg will be 4 ,200 multiplied by 5 degrees celsius.

02:41 It equals to 2100 ,000 joules.

02:47 This is cpt2...

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