Answers are: a) 2.4*10^3 kJ b) 160g glucose c) 3.7 g water per minute d) about 1

Question

Answers are: a) 2.4*10^3 kJ b) 160g glucose c) 3.7 g water per minute d) about 17 degrees Fahrenheit Mountain climbers are warned to carry warm clothes at all times, and to be especially wary when their clothes get damp. Suppose that the clothes you were wearing had absorbed 1.00 kg of water and a cold wind dried them. (a) (b) What heat loss would your body have to make up to prevent a drop in temperature? How much glucose would have to be consumed to replace that loss? (Assume that the heat produced by consumption of glucose can be obtained from the enthalpy of combustion.) (c) If your body normally puts out 150 W, what would be the maximum rate of evaporation of water (g min1) that you could tolerate without a drop in temperature? Suppose that your body did not make up the heat loss. If you weigh 60.0 kg and have the same heat capacity as water, what would be the drop in your body temperature? (d

Explanation / Answer

Mass of water absorbed = 1.00 kg

Part a

Without drop in temperature heat loss = mass of water x heat of vaporization of water

= 1 kg x 2400 kJ/kg

= 2.4 x 10^3 kJ

Part b

Enthalpy of combustion of glucose (C6H12O6)

H = (2700 kJ/Mol) x (1mol/180 g) x (1000g/1kg)

= 15000 kJ/kg

Heat loss = 2.4 x 10^3 kJ

Amount of glucose required = Heat loss / H

= (2.4 x 10^3 kJ) / (15000 kJ/kg)

= 0.160 kg x 1000g/kg = 160 g

Part c

Maximum rate of evaporation of water

= (150 J/s x 1kJ/1000J x 60s/min) / (2400 kJ/kg)

= 0.00375 kg/min x 1000g/kg

= 3.75 g/min

Part d

If heat loss did not make up

Heat loss = mass of person x Cp x drop in temperature

2400 kJ/kg = 60 kg x 4.184 kJ/kg-°C x drop in temperature

drop in temperature = 9.56°C

Drop in temperature in °F = (9.56 x 9/5) = 17.20 °F

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