A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of the water from $\mathbf{0}\mathbf{.00}\mathbf{°}\mathbf{C}\mathbf{}\mathbf{}$to the body temperature of$\mathbf{37.0}\mathbf{°}\mathit{C}$. How many liters of ice water would have to be consumed to burn off(about 1 lb) of fat, assuming that burning this much fat requires$\mathbf{3500}\mathbf{}\mathbf{\text{cal}}\mathbf{}$be transferred to the ice water? Why is it not advisable to follow this diet? ($\mathbf{\text{Oneliter}}\mathbf{}\mathbf{=}\mathbf{}{\mathbf{10}}^{\mathbf{3}}{\mathbf{\text{cm}}}^{\mathbf{\text{3}}}$. The density of water is$\mathbf{1.00}\mathbf{}{\mathbf{\text{\hspace{0.17em}g/cm}}}^{\mathbf{\text{3}}}$.)

1. Initial temperature,${\mathrm{T}}_{\mathrm{i}}=0.0°\mathrm{C}$
2. Final body temperature,${\mathrm{T}}_{\mathrm{f}}=37.0°\mathrm{C}$
3. Fat to be consumed,$\mathrm{m}=454\text{\hspace{0.17em}gor}0.454\text{\hspace{0.17em}kg}$
4. Heat required to transfer the given fat,$\mathrm{Q}=3500\text{\hspace{0.17em}calorieor}35\times {10}^5\text{cal}$
5. The density of water,$\rho =1{\text{\hspace{0.17em}\hspace{0.17em}g/cm}}^{\text{3}}$
6. Specific heat of water,$\mathrm{c}=1{\text{\hspace{0.17em}\hspace{0.17em}cal/g}}^{\text{0}}\text{C}$

A substance's specific heat capacity is the amount of energy required to raise its temperature by one degree Celsius. According to the nutritionist, a Calorie of burn is equivalent to 1000cal energy burned. Hence, using this value to get the exact energy required to be burnt and substituting it in the equation of heat transferred or produced will give the amount of water to be absorbed.

Formula:

The heat energy transferred by a body,$\mathrm{Q}=\mathrm{mc\Delta T}$ …(i)

Where, $\mathrm{m}$= mass

$\mathrm{c}$= specific heat capacity

$\mathrm{\Delta T}$= change in temperature

$\mathrm{Q}$= required heat energy

The volume of a body in terms of density, $V=\frac{m}{\rho }$

Using the formula of equation (i), the mass of the water to be consumed is given as:

$\begin{array}{c}\mathrm{m}=\frac{\mathrm{Q}}{\mathrm{c\Delta T}}\\ =\left[\frac{35\times {10}^5\text{\hspace{0.17em}cal}}{1{\text{\hspace{0.17em}cal/g}}^{\text{0}}\text{C}\times (37-0)°\mathrm{C}}\right]\mathrm{g}\\ =94.6\times {10}^3\mathrm{g}\end{array}$

Again, the density of the water is given as:

$\begin{array}{c}\mathrm{\rho }=1{\text{\hspace{0.17em}\hspace{0.17em}g/cm}}^{\text{3}}\\ =\frac{1\mathrm{g}}{1/1000\text{\hspace{0.17em}\hspace{0.17em}liter}}\text{(}\because {\text{1liter=1000cm}}^3\text{)}\\ =1000\mathrm{g}/\text{liter}\end{array}$

So, the required amount or the volume of the water to be consumed is given by equation (ii) as:

$\begin{array}{c}\mathrm{V}=\frac{94.6\times {10}^3\text{g}}{1000\text{\hspace{0.17em}g}/\text{liter}}\\ =94.6\text{\hspace{0.17em}liter}\end{array}$

Hence, the person needs to consume$94.6\text{\hspace{0.17em}liter}$ of water.