Question

(a) A person drinks four glasses of cold water \(\left(3.0^{\circ} \mathrm{C}\right)\) every day. The volume of each glass is \(2.5 \times 10^{2} \mathrm{~mL}\). How much heat (in kJ) does the body have to supply to raise the temperature of the water to \(37^{\circ} \mathrm{C},\) the body temperature? (b) How much heat would your body lose if you were to ingest \(8.0 \times 10^{2} \mathrm{~g}\) of snow at \(0^{\circ} \mathrm{C}\) to quench your thirst? (The amount of heat necessary to melt snow is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).)

Short answer

142.12 kJ to heat the water; 267.04 kJ lost melting the snow.

Step-by-step solution

Calculate the Heat Required for One Glass of Water

The formula to calculate the heat energy required to raise the temperature is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is the specific heat capacity of water (\( 4.18 \, \text{J/g}^\circ\text{C} \)), and \( \Delta T \) is the temperature change. First, convert the volume of one glass to mass: \( 250 \, \text{mL} = 250 \, \text{g} \) since the density of water is \( 1 \, \text{g/mL} \). The temperature change, \( \Delta T \), is \( 37^\circ \text{C} - 3^\circ \text{C} = 34^\circ \text{C} \). Finally, plug these into the formula: \( q = 250 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 34^\circ \text{C} \).

Convert Joules to Kilojoules for One Glass

Calculate \( q \) from Step 1: \( q = 250 \, \text{g} \times 4.18 \, \text{J/g}^\circ\text{C} \times 34^\circ\text{C} = 35530 \, \text{J} \). To convert to kilojoules, divide by \( 1000 \): \( 35530 \, \text{J} = 35.53 \, \text{kJ} \).

Calculate Total Heat for Four Glasses

Since the person drinks 4 glasses, multiply the heat for one glass by 4: \( 35.53 \, \text{kJ} \times 4 = 142.12 \, \text{kJ} \).

Calculate the Heat Loss by Melting the Snow

First, find the number of moles of snow: \( 8.0 \times 10^2 \, \text{g} \div 18 \, \text{g/mol} = 44.44 \text{ moles} \). Multiply the number of moles by the heat of fusion: \( 44.44 \text{ moles} \times 6.01 \, \text{kJ/mol} = 267.04 \, \text{kJ} \).

Conclude Total Heat Calculations

The body supplies \( 142.12 \, \text{kJ} \) to raise the water temperature to body temperature, and it would lose \( 267.04 \, \text{kJ} \) by ingesting \( 800 \, \text{g} \) of snow.

Key concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in calorimetry, and it defines the amount of heat needed to change a unit mass of a substance by one degree Celsius. For water, this value is particularly significant because of its relatively high specific heat capacity of 4.18 J/g°C. This means that water can absorb a lot of heat energy without a significant change in temperature. Such a property is why water is often used in heating and cooling applications.

When you want to calculate how much heat is needed to change the temperature of a certain mass of water, you use the formula:

• \[ q = mc\Delta T \]

Here,

• \( q \) is the heat added in Joules,
• \( m \) is the mass of the water in grams,
• \( c \) is the specific heat capacity, and
• \( \Delta T \) is the change in temperature in degrees Celsius.

Understanding specific heat capacity allows us to predict how much energy is involved during temperature changes, a useful tool in various scientific and engineering tasks.

Heat of Fusion

The heat of fusion is the amount of heat energy required to change a substance from a solid to a liquid at its melting point, without changing its temperature. This concept comes into play when you're thinking about melting ice, which requires energy even though the temperature remains constant. For water, the heat of fusion is 6.01 kJ/mol.

Consider when snow melts: although the snow is at its melting point (0°C), energy is still required for the phase change from solid to liquid. This energy, in the context of our exercise, is calculated based on the number of moles of snow and the heat of fusion:

• Calculate the number of moles of water in the snow using its mass and the molar mass of water (18 g/mol).
• Multiply the number of moles by the heat of fusion value to find the total heat required.

This principle is essential in understanding energy transformations during phase changes in nature and industry.

Temperature Change

Temperature change is a core concept of calorimetry, representing the difference between the initial and final temperatures of a substance undergoing a heat exchange. It is denoted as \( \Delta T \) and plays a key role in calculating heat energy using the specific heat capacity formula.

In the given exercise, the initial temperature of the drinking water is 3°C, and it needs to reach the body temperature of 37°C. The change in temperature, \( \Delta T \), therefore, equals 34°C.

Understanding temperature change is critical for performing calculations in calorimetry. It determines how much energy an object gains or loses during a thermal process. This concept is applicable in numerous real-world situations, like heating systems, cooking, and even weather predictions where heat transmission is involved.

Mass to Energy Conversion

Mass to energy conversion in calorimetry involves calculating how much energy corresponds to a given mass undergoing a temperature or phase change. Unlike the famous E=mc² formula of relativity, this conversion in calorimetry uses equations like \( q = mc\Delta T \) and is more about energy transformation.

In calorimetry, it's not just about the mass alone, but how that mass interacts with temperature changes or phase transitions.

• When raising the temperature of a given mass, as in heating water, you use the specific heat capacity.
• When melting snow, you use the heat of fusion to understand how much energy is needed for the phase change per gram or mole of substance.

Understanding these conversions helps you calculate the total energy change in practical applications like heating elements and cooling systems, making it essential for both scientific studies and daily problem-solving.