Question

Evaporating sweat cools the body because evaporation is an endothermic process: $$\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=+44.01 \mathrm{~kJ}$$ Estimate the mass of water that must evaporate from the skin to cool the body by \(0.50^{\circ} \mathrm{C}\). Assume a body mass of \(95 \mathrm{~kg}\) and assume that the specific heat capacity of the body is \(4.0 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).

Short answer

The mass of water that needs to evaporate to cool the body by 0.50°C is ~1902 g.

Step-by-step solution

Calculate the Amount of Heat Required to Cool the Body

First, determine the amount of heat needed to cool a 95 kg body by 0.50°C using the specific heat capacity formula, which is: q = mcΔT. Here, m = 95000g (since 1 kg = 1000g), c = the specific heat capacity which is 4.0 J/g°C, and ΔT = the temperature change, which here is 0.50°C. Now calculate q: q = 95000g * 4.0 J/g°C * 0.50°C.

Convert the Heat Energy Required to Mass of Water

Using the enthalpy change for the evaporation process, calculate the mass of water that must evaporate. Since we have q (heat energy required to cool the body), we use that the enthalpy change is +44.01 kJ per mole of water. First, convert q from J to kJ (since 1 kJ = 1000 J). Then, to find the mass of water evaporated, set up a proportion: if 44.01 kJ of energy is required to evaporate 1 mole of water (18 g), then q kJ will evaporate how many grams? Solve for the mass of water.

Solve for the Mass of Water Evaporated

Expressing q from Step 1 in kJ, and setting up the proportion from Step 2, we have: (q kJ / 44.01 kJ) = (mass of H2O / 18 g). Solve for the mass of H2O using cross-multiplication.

Key concepts

Specific Heat Capacity

Specific heat capacity is a measure of the amount of heat energy required to raise the temperature of a given mass of a substance by one degree Celsius. Essentially, it indicates how much heat energy is needed to heat up a substance and also gives an idea of how a substance can absorb and hold heat. This property varies across different materials, making it a crucial factor in understanding and performing heat energy calculations.

For instance, in the given exercise, the specific heat capacity of the body is used to calculate the amount of heat energy needed to cool the body by a specific temperature change. The formula to calculate the heat (q) is: \[q = mc\Delta T\] where \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. This calculation is foundational in determining how much heat is lost or gained during processes like the endothermic evaporation of sweat from the skin.

Enthalpy Change

Enthalpy change, denoted as \(\Delta H\), represents the amount of heat absorbed or released by a system at constant pressure during a chemical or physical process. It's a crucial concept in thermodynamics and is commonly expressed in units of kilojoules per mole (kJ/mol). The sign of the enthalpy change indicates the nature of the process: positive for endothermic (heat absorbed) and negative for exothermic (heat released).

In the context of the exercise, the enthalpy change of +44.01 kJ/mol corresponds to the energy required for the phase change from liquid water to gaseous water, signifying that evaporation is an endothermic process. This value is used to determine the mass of water that must evaporate from the body to achieve the desired cooling effect.

Heat Energy Calculations

Heat energy calculations involve determining the amount of heat gained or lost during a process. These calculations require a sound understanding of various physical properties, such as specific heat capacity and enthalpy change, as well as mastery in applying relevant formulas.

In the exercise, the heat energy calculation is performed in two steps: First, by calculating the heat required to cool the human body with the given specific heat capacity, and second, by converting this heat energy into the mass of water that needs to evaporate using the enthalpy change of vaporization. Heat energy (q) is initially calculated in Joules but is then converted into kilojoules to match the enthalpy unit, allowing for the subsequent proportion calculation to determine the mass of water that evaporates.

Molar Mass of Water

The molar mass of a substance is the mass of one mole of that substance. It's a critical constant in stoichiometry as it bridges the gap between the microscopic scale of atoms and molecules and the macroscopic world of grams and kilograms. For water (H2O), the molar mass is approximately 18 grams per mole.

This value is significant in our exercise because it allows us to convert between moles and grams. Once the amount of heat energy required to cool the body is calculated and converted to kilojoules, this value is then used alongside the molar mass of water and the enthalpy change to find the equivalent mass of water that must evaporate to achieve the cooling. This conversion is essential to understand how much matter is involved in the physical process of the body's cooling via evaporation of sweat.