A hiker caught in a rainstorm absorbs 1.00 L of water in her clothing. If it is windy so that the water evaporates quickly at 20°C, how much heat is required for this process? b. If all this heat is removed from the hiker (no significant heat was generated by metabolism during this time), what drop in body temperature would the hiker experience? The clothed hiker weighs 60 kg, and you can approximate the heat capacity of hiker and clothes as equal to that of water. ( Moral : stay out of the wind if you get your clothes wet.) c. How many grams of sucrose would the hiker have to metabolize (quickly) to replace the heat of evaporating 1.00 L of water so that her temperature would not change? You can use the enthalpy of reaction at 25°C; the sucrose ( s ) reacts with oxygen ( g ) to give carbon dioxide ( g ) and water ( l ).

Respuesta :

Answer:

(A) Q = 2.26×10⁶J

(B) ΔT = 9°C

(C)

Explanation:

We have been given the mass of the hiker, the volume of water from which we can calculate the mass knowing that the density if water is 1000kg/m³.

Evaporation is a phase change and occurs at a constant temperature. We would use the latent heat of vaporization to calculate the amount of heat evaporated.

We would then equate this to the heat change it brings about in the hiker's body and then calculate the temperature drop.

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