Alternative diets

Classical Mechanics Level 2

Paul wants to lose weight, but he doesn't want to do sports or change his eating. He thinks about three different diets with which he can still lose weight:

  • Cold Drinks Diet: Instead of water at room temperature ( 2 0 ∘ C 20^\circ \,\text{C} ) Paul drinks only ice-cooled water. Paul drinks 2 liters every day.
  • Cold Room Diet: Paul reduces the temperature of his living room about 5 degrees Celsius without changing his clothes. At normal room temperature and sitting, his body releases about 80 watts of heat. Paul is in the living room for 4 hours a day, watching television.
  • Horror Movie Diet: Paul watches a horror movie of 90 minutes length every day. We assume, that he burns twice as many calories as he does with his usual television program.

With which diet does Paul burn the most calories?

Hints: A calorie equals 4.187 Joules and corresponds to the required amount of energy to heat 1 gram of water about 1 degree Celsius.

Freezing Diet Horror Movie Diet Cold Drinks Diet

Markus Michelmann by Markus Michelmann , 35, Germany

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1 solution

  • Cold Drinks Diet: In order to heat the drinking water up to body temperature, the body needs the energy Q = C m ( T body − T water ) Q = C m (T_\text{body} - T_\text{water}) With the mass m = 2 kg m = 2 \,\text{kg} of 2 liters and the heat capactiy C = 1 kcal kg K C = 1 \frac{\text{kcal}}{\text{kg}\,\text{K}} the additional energy demand results Δ Q = C m Δ T water = 40 kcal \Delta Q = C m \Delta T_\text{water} = 40\, \text{kcal}
  • Cold Room Diet: The heat loss of the body is proportional to the temperature difference between the body and the room Q ˙ = k A ( T body − T room ) \dot Q = k A (T_\text{body} - T_\text{room}) with the heat conductivity k k of skin and clothes, the body surface A A and the body temperature T body = 3 7 ∘ C T_\text{body} = 37^\circ \,\text{C} . The 5 degree reduction of the room temperature increases the heat flow to Q ˙ = 37 − 15 37 − 20 80 W ≈ 103.5 W \dot Q = \frac{37 - 15}{37 - 20} 80 \,\text{W} \approx 103.5 \, \text{W} Therefore, the body needs additional 23.5 watts to sustain the body temperature. In relation to the entire television evening this corresponds to an energy requirement Δ Q = 23.5 W ⋅ 4 hours = 23.5 ⋅ 4 ⋅ 60 ⋅ 60 4187 kcal ≈ 81 kcal \Delta Q = 23.5 \,\text{W} \cdot 4\,\text{hours} = \frac{23.5 \cdot 4 \cdot 60 \cdot 60}{4187} \, \text{kcal} \approx 81 \, \text{kcal}
  • Horror Movie Diet : Watching the horror film, the body loses additional 80 watts, which results into an extra energy demand of Δ Q = 80 W ⋅ 90 min = 80 ⋅ 90 ⋅ 60 4187 kcal ≈ 103 kcal \Delta Q = 80 \,\text{W} \cdot 90 \,\text{min} = \frac{80 \cdot 90 \cdot 60}{4187} \text{kcal} \approx 103\,\text{kcal}
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