Ice Diet
Task number: 1279
Water does not provide the human body with any usable energy. However, when heating some ice to body temperature, the human body burns off energy (energy needs to be supplied). How many ice cubes with a mass of 50 g and a temperature of −20 °C do you have to eat to lose five kilograms of fat? Burning one gram of fat during metabolic processes in the body releases the heat of 38 kJ.
Hint
The energy released by burning the body fat is used for
- heating the ice to the melting point,
- melting the ice,
- heating the melted water to the body temperature.
Numerical values
m1 = 50 g = 0.050 kg the mass of an ice cube tI = −20 °C the temperature of ice mf = 5 kg the mass of burnt fat H = (38 kJ)/(1 g) = 38×106Jkg−1 the energy released by burning fat N = ? the number of ice cubes Other necessary values:
tB = 36.6 °C the body temperature tm = 0 °C the melting point of ice cI = 2.1 kJkg−1K−1 = 2100 Jkg−1K−1 the specific heat capacity of ice lfus = 334 kJkg−1 =3.34×105 Jkg−1 the specific latent heat of fusion of ice cW = 4.18 kJkg−1K−1 = 4180 Jkg−1K−1 the specific heat capacity of water Analysis
First, we will determine how much energy is released when “burning” 5 g of fat. The energy released is equal to the heat received by the ice. This heat is a sum of the heat required to warm the ice to the melting temperature, the latent heat needed to melt the ice and the heat required to warm the melted water to the body temperature. Hence we can determine the mass of ice and consequently the number of cubes.
Solution
The energy released by burning fat of a mass mf:
\[E=Hm_f.\]The heat that needs to be supplied to ice of unknown mass mI to be melted and warmed up to the body temperature:
the warming of ice: \[Q_1 = c_I m_I (t_m - t_I)\]the melting of ice: \[L_I = l_{fus} m_I\]the warming of melted water: \[Q_2 = c_W m_I(t_B - t_m)\]In total:: \[Q= c_I m_I (t_m - t_I) +l_{fus} m_L + c_W m_I(t_B - t_m)=\]\[\ \, \ = m_I[c_I (t_m-t_I) + l_{fus} + c_W (t_B - t_m)].\]The energy gained by burning fat is used for warming the ice, that is to say:
\[E=Q,\]\[Hm_{f}=m_I [c_I (t_m-t_I) +l_{fus} + c_W (t_B - t_m)].\]Now we can evaluate the unknown mass of ice and then substitute the numerical values:
\[m_I = \frac{Hm_f}{c_I (t_m-t_I) + l_{fus} + c_W (t_B - t_m)}\]\[m_I = \frac{38 \cdot{10^6} \cdot 5}{2100 \cdot [0-(-20)] + 3.34 \cdot{10^5} + 4180 \cdot (36.8 - 0)}\,\mathrm kg \dot= \]\[\ \, \ \dot= 360 \,\mathrm kg \]Finally, we determine the number of ice cubes:
\[N = \frac{m_I}{m_1}=\frac{Hm_f}{m_1[c_I (t_m-t_I) + l_{fus} + c_W (t_B - t_m)]}\]\[N \dot= 7200\]Answer
When on the “ice diet”, you would have to eat about 7200 ice cubes to lose five kilograms of fat, which corresponds to about 360 kilograms of ice.
