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2 March, 09:18

A 100 kg individual consumes 1200 kcal of food energy a day. Calculate

(a) the altitude change, in m, if the food energy content was converted entirely into lifting the indi-

vidual under normal earth gravity.

(b) the velocity, in m/s, if the food energy content was converted entirely into accelerating the indi-

vidual from rest.

(c) the final temperature, in ◦C, of a 100 kg mass of liquid water initially at the normal human body

temperature and heated with the energy content of the food. You can use a liquid water specific

heat of 4.1 kJ/kg K.

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Answers (1)
  1. 2 March, 09:22
    0
    (a) 5142.86 m

    (b) 317.5 m/s

    (c) 49.3 degree C

    Explanation:

    m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

    (a) Let the altitude be h

    Q = m x g x h

    504 x 10^4 = 100 x 9.8 x h

    h = 5142.86 m

    (b) Let v be the speed

    Q = 1/2 m v^2

    504 x 10^4 = 1/2 x 100 x v^2

    v = 317.5 m/s

    (c) The temperature of normal human body, T1 = 37 degree C

    Let the final temperature is T2.

    Q = m x c x (T2 - T1)

    504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

    T2 = 49.3 degree C
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