Question

A soft-drink can contains about 0.35 kg of liquid at 5°C. Drinking this liquid can actually consume some of the fat in the body, since energy is needed to warm the liquid to body temperature (37°C). How many food Calories should the drink have so that it is in perfect balance with the heat needed to warm the liquid (essentially water)?

Short answer

The food calories contained in the drink are $11.2\mathrm{k}\mathrm{c}\mathrm{a}\mathrm{l}$.

Step-by-step solution

Characteristics of specific heat

Objects with a lower value of specific heat have a faster temperature increase and decrease. They are very sensitive to temperature changes.

Given data

The mass of the liquid is $m=0.35\mathrm{k}\mathrm{g}$.

The initial temperature is $T_1=5\circ \mathrm{C}$.

The final temperature is $T_2=37\circ \mathrm{C}$.

Evaluation of the heat required to warm the liquid

The heat required to warm the liquid can be calculated as:

$Q=m{c}_{\mathrm{w}}\left(T_2-T_1\right)$

Here, $c_{\mathrm{w}}$ is the specific heat of the water, and its value is $$\begin{gathered} 4186\mathrm{J}/\phantom{\mathrm{J}\mathrm{k}\mathrm{g}\cdot \circ \mathrm{C}} \\ \mathrm{k}\mathrm{g}\cdot \circ \mathrm{C} \end{gathered}$$.

Substitute the values in the above equation.

$$\begin{gathered} \begin{array}{c}Q=\left(0.35\mathrm{k}\mathrm{g}\right)\left(4186\mathrm{J}/\phantom{\mathrm{J}\mathrm{k}\mathrm{g}\cdot \circ \mathrm{C}} \\ \mathrm{k}\mathrm{g}\cdot \circ \mathrm{C}\right)\left(37\circ \mathrm{C}-5\circ \mathrm{C}\right)\\ =\left(46883.2\mathrm{J}\right)\left(\frac{2.39\times 10-4\mathrm{k}\mathrm{c}\mathrm{a}\mathrm{l}}{1\mathrm{J}}\right)\\ =11.2\mathrm{k}\mathrm{c}\mathrm{a}\mathrm{l}\end{array} \end{gathered}$$

Thus, the required food calories are $11.2\mathrm{k}\mathrm{c}\mathrm{a}\mathrm{l}$.