The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. a. What mass of water m

Question

The evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals.

a. What mass of water must evaporate from the skin of a 66.0 kg man to cool his body 1.30 °C? The heat of vaporization of water at body temperature (37.0 ∘C) is 2.42×10^6J/kg. The specific heat capacity of a typical human body is 3480 J/(kg⋅K).
b. What volume of water must the man drink to replenish the evaporated water? Compare this result with the volume of a soft-drink can, which is 355 cm^3

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Latifah 4 years 2021-08-18T18:18:38+00:00 1 Answers 9 views 0

Answers ( )

  1. Sigrid
    0
    2021-08-18T18:19:49+00:00
    Reply

    Answer:

    a) m = 26.03 kg

    b) V = 26000 cm³  

    Explanation:

    a) The mass of water that should evaporate from the skin can be calculated using the following equation:

    m*Q = M*C*\Delta T                        

    Where:                                

    m: is the mass of water =?

    Q: is the heat of vaporization of water = 2.42×10⁶ J/kg      

    C: is the specific heat capacity = 3480 J/(kg*K)

    ΔT: is the temperature difference = 1.30 °C = 274.3 K  

    M: is the mass of the man = 66.0 kg

    m= \frac{M*C*\Delta T}{H} = \frac{66.0 kg*3480 J/(kg*K)*274.3 K}{2.42\cdot 10^{6} J/kg} = 26.03 kg              

    Hence, 26.03 kg of the mass of water must evaporate from the skin.  

    b) The volume (V) of water is:    

    V = \frac{m}{d}

    Where:

    d: is the density of water = 997 kg/m³          

    V = \frac{26.03 kg}{997 kg/m^{3}} = 0.026 m^{3} = 26000 cm^{3}  

    Compared to a soft drink can of 355 cm³, the man should drink approximately 73 cans of soft drink to compensate for the evaporated water.  

    I hope it helps you!

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