Question

When glucose, a sugar, reacts fully with oxygen, carbon dioxide and water are produced:

${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{12}}{\mathbf{O}}_{\mathbf{6}}\left(\mathbf{s}\right)\mathbf{+}\mathbf{6}{\mathbf{O}}_{\mathbf{2}}\left(\mathbf{g}\right)\to \mathbf{6}{\mathbf{CO}}_{\mathbf{2}}\left(\mathbf{g}\right)\mathbf{+}\mathbf{6}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(\mathbf{l}\right){\mathbf{\Delta H}}^{\mathbf{o}}\mathbf{=}\mathbf{-}\mathbf{2820}\mathbf{kJ}$

Suppose a person weighing 50 kg (mostly water, with specific heat capacity $\mathbf{4}\mathbf{.}\mathbf{18}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$) eats a candy bar containing 14.3 g glucose. If all the glucose reacted with oxygen and the heat produced were used entirely to increase the person’s body temperature, what temperature increase would result? (In fact, most of the heat produced is lost to the surroundings before such a temperature increase occurs.)

Short answer

The temperature is increased by 192.9 K.

Step-by-step solution

Given Information.

The weight of the person is 50 kg.

The amount of glucose is 14.3 g.

The value of c is $4.18{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

The value of ΔH°s -2820 kJ.

Specific heat.

The quantity of heat required to elevate the temperature of one gram of a particular substance by one degree celsius is termed specific heat. The unit is indicated in calories.

The temperature that increases in the reaction.

The heat can be given as:

$\begin{array}{l}\mathrm{Heat}=\mathrm{Amountofsubstance}\times \Delta \mathrm{H}°\\ =14.3\mathrm{g}\times 2820\times {10}^3\mathrm{J}\\ =40326\times {10}^3\mathrm{J}\end{array}$

The change in temperature can be given as:

$\begin{array}{l}\mathrm{Q}=\mathrm{mc\Delta T}\\ \mathrm{\Delta T}=\frac{\mathrm{Q}}{\mathrm{mc}}\\ =\frac{40326\times {10}^3\mathrm{J}}{50\times {10}^3\mathrm{g}\times 4.18{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}}\\ =192.9\mathrm{K}\end{array}$

Hence, the temperature is increased by 192.9 K.