Problem 54 - An Introduction to Thermal Physics

Question

A 60-kg hiker wishes to climb to the summit of Mt. Ogden, an ascent of 5000 vertical feet \((1500 \mathrm{m})\) (a) Assuming that she is \(25 \%\) efficient at converting chemical energy from food into mechanical work, and that essentially all the mechanical work is used to climb vertically, roughly how many bowls of corn flakes (standard serving size 1 ounce, 100 kilocalories) should the hiker eat before setting out? (b) As the hiker climbs the mountain, three-quarters of the energy from the corn flakes is converted to thermal energy. If there were no way to dissipate this energy, by how many degrees would her body temperature increase? (c) In fact, the extra energy does not warm the hiker's body significantly; instead, it goes (mostly) into evaporating water from her skin. How many liters of water should she drink during the hike to replace the lost fluids? (At \(25^{\circ} \mathrm{C},\) a reasonable temperature to assume, the latent heat of vaporization of water is \(580 \mathrm{cal} / \mathrm{g}, 8 \%\) more than at \(100^{\circ} \mathrm{C} .\) )

Step-by-Step Solution

Answer

(a) 9 bowls; (b) Temperature increase 10.54°C; (c) Needs 1.09 liters of water.

Step 1: Calculate Mechanical Work Required

Determine the gravitational potential energy required for the hiker to ascend 1500 meters. This can be calculated using the formula for gravitational potential energy: \[ \text{Work} = m \cdot g \cdot h \] where \( m = 60 \text{ kg} \) is the mass of the hiker, \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity, and \( h = 1500 \text{ m} \) is the height. Calculate: \[ \text{Work} = 60 \times 9.8 \times 1500 = 882000 \text{ J} \]

Step 2: Adjust for Efficiency

Convert the required mechanical work into chemical energy, considering the hiker's efficiency as \(25\%\). Since she is only \(25\%\) efficient, the chemical energy required is \(4\) times the mechanical energy. Thus, the required energy is \[ \text{Energy} = \frac{882000 \text{ J}}{0.25} = 3528000 \text{ J} \]

Step 3: Convert Energy to Kilocalories

Convert the energy from joules to kilocalories, where \(1 \text{ kcal} = 4184 \text{ J}\). Thus, \[ \text{Energy in kcal} = \frac{3528000}{4184} \approx 843 \text{ kcal} \]

Step 4: Calculate Number of Bowls of Corn Flakes

Given that each bowl (1 ounce serving) of corn flakes provides \(100\) kilocalories, calculate the number of bowls needed: \[ \text{Number of bowls} = \frac{843 \text{ kcal}}{100 \text{ kcal/bowl}} = 8.43 \] Thus, she should eat approximately \(9\) bowls to have enough energy.

Step 5: Calculate Temperature Increase

Determine the thermal energy generated, which is \(75\%\) of the total energy consumed. Use the formula \( Q = mc\Delta T \) to find the temperature increase, where \( m = 60 \times 1000 = 60000 \text{ g} \), \( c \) is specific heat capacity of water \( = 1 \text{ cal/g°C} \), and \( Q = 843 \times 0.75 \times 1000 = 632250 \text{ cal} \). Thus, \[ \Delta T = \frac{Q}{mc} = \frac{632250}{60000} \approx 10.54 \text{ °C} \]

Step 6: Calculate Water Loss for Cooling

Using the latent heat of vaporization of water at \(25^{\circ} \text{C} = 580 \text{ cal/g} \), calculate the mass of the water evaporated: \[ m = \frac{632250 \text{ cal}}{580 \text{ cal/g}} \approx 1090 \text{ g} \] and convert this to liters, knowing density of water is \(1 \text{ kg/L}\): \[ \text{Liters of water} = 1.09 \text{ L} \]

Key Concepts

Gravitational Potential Energy

When climbing a mountain, like the 1500-meter vertical ascent of Mt. Ogden, a hiker must exert energy to raise her body against Earth's gravitational force. This is known as gravitational potential energy. It can be calculated using the formula:
\[ \text{Work} = m \cdot g \cdot h \] where:

• \( m \) is the mass of the hiker (60 kg).
• \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \)).
• \( h \) is the height of the climb (1500 m).

Multiplying these values, we find that the hiker needs to perform \( 882000 \text{ J} \) of work to reach the top. This energy is used entirely to elevate her body mass upwards against gravity, demonstrating the core concept of gravitational potential energy in action.
Understanding this helps in realizing how much energy is converted from our food just to gain vertical distance on a hike.

Chemical to Mechanical Energy Conversion Efficiency

The process of converting the chemical energy stored in food into mechanical energy for climbing involves efficiency considerations. In the context of our hiker, the efficiency is given as 25%. This means that only a quarter of the chemical energy consumed becomes useful mechanical work; the rest is lost, mainly as heat. To find out how much energy the hiker requires, we must first calculate the total chemical energy needed given her efficiency.The formula to determine the chemical energy needed is:\[ \text{Energy} = \frac{\text{Mechanical Work}}{\text{Efficiency}} \] Given that the mechanical work needed is \( 882000 \text{ J} \), and the efficiency is 25% (or 0.25 as a decimal), the calculation becomes:\[ \text{Energy} = \frac{882000}{0.25} = 3528000 \text{ J} \] This calculation highlights how critical understanding efficiency is when planning for physical activities, emphasizing the significant energy requirements due to inefficiencies in biological energy conversion.

Specific Heat Capacity

Specific heat capacity is a property that indicates how much energy is needed to change the temperature of a substance. It plays a crucial role in understanding how a hiker's body temperature would rise if excess energy were not dissipated. For human bodies, which are largely water, we use the specific heat capacity of water, given as \( 1 \text{ cal/g°C} \).The rise in body temperature, \( \Delta T \), can be found using the relation:\[ Q = mc\Delta T \] where:

• \( Q \) is the thermal energy in calories. For the hiker, this is \( 843 \times 0.75 \times 1000 = 632250 \text{ cal} \).
• \( m \) is the mass of the hiker in grams (60000 g, since 60 kg).
• \( c \) is the specific heat capacity.

Reorganizing to find \( \Delta T \):\[ \Delta T = \frac{632250}{60000} \approx 10.54 \text{ °C} \] Understanding specific heat capacity is vital for appreciating how much thermal energy the body can absorb before a significant temperature increase occurs.

Latent Heat of Vaporization

During vigorous activities, our bodies often cool down by evaporating sweat. This process uses latent heat of vaporization, a concept that indicates the energy required to convert water from liquid to vapor without changing its temperature. At 25°C, this value for water is 580 cal/g.When the hiker climbs the mountain, the excess energy is used to evaporate sweat, effectively cooling her down. To determine the amount of water evaporated, the following relation is used:\[ m = \frac{Q}{L} \] where:

• \( Q \) is the energy used in evaporation (632250 cal).
• \( L \) is the latent heat of vaporization (580 cal/g).

The mass of water evaporated, \( m \), is given by:\[ m = \frac{632250}{580} \approx 1090 \text{ g} \] Converting grams to liters, given that \( 1 \text{ kg/L} \), results in approximately \( 1.09 \text{ L} \). This means the hiker should drink about one liter of water to replace the fluid lost through sweat. Understanding this concept helps in hydration planning, ensuring hiker’s endurance during long climbs.