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Thermodynamics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsThermodynamics5 Marks
Question
A person of mass $60kg$ wants to lose $5kg$ by going up and down a $10m$ high stairs. Assume he burns twice as much fat while going up than coming down. If $1kg$ of fat is burnt on expending $7000$ kilo calories, how many times must he go up and down to reduce his weight by $5kg$?
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Answer

Gravitational potential energy (PE) of an object at height (h) is mgh. The energy losses by person in the form of fat will be utilised to increase PE of the person. As it is given that he burns twice as much fat while going up than coming down. Thus, the calorie consumed by the person in going up is mgh, and calorie consumed by the person in comming down is 1/ 2 mgh According to the problem, height of the stairs $= h = 10 m$ Work done to burn $5kg$ of fat $= (5kg) (7000 \times 10^3 cal) (4.2J/ cal) = 147 \times 10^6J$ Work done towards burning of fat in one trip (up and down the stairs )$=\text{mgh}+\frac{1}{2}\text{mgh}=\frac{3}{2}\text{mgh}$
$=\frac{3}{2}(60\text{kg})(10\text{m/ s}^2)(10m)=9\times10^3\text{J}$
(as only half the work done while coming down is useful in burning fat ) $\therefore$ Number of times, the person has to go up and down the stairs (no. of trips required)
$\text{N}=\frac{147\times10^6\text{J}}{9\times10^3\text{J}}=16.3\times10^3\text{times}$

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