You read in a popular diet blog that the author claims you can lose up to half a pound of body fat per week if, instead of drinking 1.7 liters (i.e., 1.7 kilograms) of water every day, you can eat the same amount of water in the form of ice. The reasoning is that your body must first melt the ice and then bring the ice water up to the body temperature of 37°C, which is a process that requires a significant amount of energy at the expense of "burned" body fat. Thankfully, you just finished discussing heat and phase changes in your physics class, so you can calculate exactly how much energy is expended. The specific heat and latent heat of fusion for water are 4.186 x 10^3 J/(kg:°C) and 3.34x10^9 J/kg, respectively.

(a) How many kilocalories do you expend when you eat 1.7 kg of ice at 0°C? Note that 1 kcal = 4186 J.
(b) Assuming the metabolism of one pound of body fat produces 3500 kcal of energy, how many pounds of body fat could you lose in a week just by eating 1.7 kilograms of ice every day?

Respuesta :

Answer:

a)  Q = 8.31 10⁵ J    b)  M = 0.39 lb

Explanation:

there is a change in temperature and the other where there are changes in state.  

    Q = m [tex]c_{e}[/tex] ΔT = m [tex]c_{e}[/tex] ([tex]T_{f}[/tex] –T₀)  

    Q = m L  

a) let's look for the heat to melt the ice  

Latent heat of ice melting 3.34 105 J / kg  

    Q₁ = m L  

    Q₁ = 1.7 3.34 10⁵  

   Q₁ = 5.678 10⁵ J

 

water to 37°C

   Q₂ = m [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-T₀)  

   Q₂ = 1.7 4.186 10³ (37 - 0)  

   Q₂ = 2,632 10⁵ J  

The total heat in the process is  

   Q = Q₁ + Q₂  

   Q = (5,678 + 2,632) 10⁵ J  

   Q = 8.31 10⁵ J  

b) say that 1lb of fat produces

       3500 kcal (4186 J / 1kcal) = 1,465 10⁷ J  

Let's use a rule of proportions, if in one day 8.31 10⁵ J is needed to process the ice how much is needed in 7 days  

       Qt = 7 8.31 10⁵ J  

       Qt = 58.17 10⁵ J  

Amount of fat  

      M = 1lb (58.17 10⁵ / 1.465 10⁷)  

      M = 39.70 10⁻² lb  

      M = 0.39 lb