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One method of reducing dangerous high body temperature very quickly is to give a patient an
alcohol rub. How many grams of alcohol must be evaporated from the surface of a 70-kg person to
reduced his temperature by 1.5 0C ? (Latent heat of vaporization of alcohol =204 cal/g)

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To calculate the amount of alcohol that must be evaporated from the surface of the person to reduce their temperature by 1.5°C, we can use the formula:
Q = mcΔT
Where:
Q = heat lost by the person (in joules)
m = mass of the alcohol (in grams) to be evaporated
c = specific heat capacity of alcohol (we assume it is similar to the specific heat capacity of water, which is approximately 4186 J/kg°C)
ΔT = change in temperature (in °C)
First, we need to convert the latent heat of vaporization of alcohol from calories to joules:
204 cal/g * 4.186 J/cal = 854.184 J/g
Now, we can calculate the heat lost by the person:
Q = mcΔT
Q = m * 4186 J/kg°C * 1.5°C
Q = 6279 m J
Since the heat lost from the person is equal to the heat needed to evaporate the alcohol, we have:
6279 m J = m * 854.184 J/g
m = 6279 J / 854.184 J/g
m ≈ 7.35 g
Therefore, approximately 7.35 grams of alcohol must be evaporated from the surface of the person to reduce their temperature by 1.5°C.
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