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Chapter 12: Problem 50

When you drink cold water, your body must expend metabolic energy in order to maintain normal body temperature \((37^{\circ} {C})\) by warming up the water in your stomach. Could drinking ice water, then, substitute for exercise as a way to 鈥渂urn calories?鈥� Suppose you expend 430 kilocalories during a brisk hour-long walk. How many liters of ice water \((0^{\circ} {C})\)) would you have to drink in order to use up 430 kilocalories of metabolic energy? For comparison, the stomach can hold about 1 liter.

Short Answer

Expert verified
To burn 430 kcal, you'd need to drink about 11.62 liters of ice water; impractical due to stomach size limits.

Step by step solution

01

Understanding the concept of heat energy

When you drink cold water, your body has to bring the water up to its normal body temperature, which costs energy. This energy expenditure is what we're calculating and comparing to the calories burned during exercise.
02

Calculating the energy required to heat water

The amount of energy required to heat water is given by the equation: \[Q = mc\Delta T\]where \(Q\) is the heat energy in calories, \(m\) is the mass of the water in grams, \(c\) is the specific heat capacity of water \((1\text{ cal/g}^{\circ}\text{C})\), and \(\Delta T\) is the change in temperature. For water from \(0^{\circ} {C}\) to \(37^{\circ} {C}, \Delta T = 37^{\circ} {C}.\)
03

Calculating the energy for 1 liter of water

First, determine the mass of 1 liter of water. Liquid water has a density of \(1\text{ g/cm}^3\), so 1 liter (which is 1000 milliliters) is equivalent to 1000 grams. Using the specific heat formula:\[Q = 1000 \text{ g} \times 1 \text{ cal/g}^{\circ}\text{C} \times 37^{\circ}\text{C} = 37000 \text{ cal} = 37\text{ kcal}\]
04

Calculating the total energy expenditure

We know from the problem that you want to expend 430 kilocalories of energy. Since each liter of ice water requires 37 kcal to heat up, find out how many liters are needed using the equation:\[\text{Total energy required} = \text{Liters of water} \times 37 \text{ kcal}\]where:\[430 \text{ kcal} = \text{Liters of water} \times 37 \text{ kcal/liter}\]Solving for Liters of water yields:\[\text{Liters of water} = \frac{430 \text{ kcal}}{37 \text{ kcal/liter}} \approx 11.62 \]
05

Comparing to stomach capacity

The stomach can hold about 1 liter. Therefore, to burn 430 kilocalories by drinking the calculated amount of ice water, you would need to drink approximately 11.62 liters, which exceeds stomach capacity by more than 11 times.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Metabolic Energy
When we talk about metabolic energy, we're referring to the energy our body uses to carry out its everyday functions, such as maintaining body temperature and digestive processes. This energy is measured in calories. Each action we do, from exercising to simply digesting food, requires a certain amount of metabolic energy.
When you drink cold water, your body must warm it to reach the normal body temperature of around 37掳C. This warming process consumes metabolic energy.
Your body doesn't distinguish between energy used for warming the water and energy used during physical exercises. All are part of your daily metabolic energy expenditure.
  • This is crucial because it shows how every calorie we use involves a balance between metabolic functions and physical activities.
  • This balance is what determines whether we gain, lose, or maintain our weight.
Heat Energy Calculation
Calculating heat energy involves understanding the basic formula: \[Q = mc\Delta T\] where:
  • \(Q\) is the heat energy required in calories,
  • \(m\) is the mass of the water (in grams),
  • \(c\) is the specific heat capacity of water, which is \(1 \, \text{cal/g掳C}\),
  • \(\Delta T\) is the change in temperature.
For instance, to calculate how much energy is required to heat 1 liter of ice-cold water (0掳C) to body temperature (37掳C), you need to know that 1 liter of water is equivalent to 1000 grams, and the temperature rise required is 37掳C.
Plug these values into our formula: \[Q = 1000 \, \text{g} \times 1 \, \text{cal/g掳C} \times 37掳C = 37000 \, \text{cal} = 37 \, \text{kcal}\]
  • Thus, warming up 1 liter of ice water in your body consumes 37 kilocalories of metabolic energy.
Specific Heat Capacity
The concept of specific heat capacity is vital to understanding energy transfer. Specific heat capacity refers to the amount of heat energy required to change the temperature of a unit mass of a substance by 1 degree Celsius. For water, this value is relatively high at \(1 \, \text{cal/g掳C}\).
This means water can absorb a large amount of heat before its temperature changes significantly. This property of water is crucial for many biological and physical processes.
  • In our body's context, when we drink cold beverages, the specific heat capacity of water determines how much metabolic energy our body must expend to raise the water from 0掳C to 37掳C.
  • Understanding this helps in comprehending how different liquids and foods can affect our metabolic rates and calorie expenditure.
Energy Expenditure Compared to Exercise
Comparing energy expenditure when heating cold water versus exercising can provide insights into effective weight management strategies. In our example, walking briskly for one hour burns approximately 430 kilocalories, the same amount needed to heat about 11.62 liters of ice water to body temperature.
However, to burn the equivalent calories by drinking cold water, one must consume more than 11 liters, far exceeding an average stomach's capacity.
  • This disparity shows that while drinking ice water does use up some calories, the amounts required to match typical exercise routines make it impractical as a sole calorie-burning strategy.
  • Regular exercise not only burns calories but also strengthens muscles and improves cardiovascular health, offering broader benefits than merely adjusting calorie expenditure via water consumption.

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Most popular questions from this chapter

Occasionally, huge icebergs are found floating on the ocean's currents. Suppose one such iceberg is 120 \({km}\) long, 35 \({km}\) wide, and 230 \({m}\) thick. (a) How much heat would be required to melt this iceberg (assumed to be at \(0^{\circ} {C}\) ) into liquid water at \(0^{\circ} {C}\) ? The density of ice is 917 \({kg} / {m}^{3} .\) (b) The annual energy consumption by the United States is about \(1.1 \times 10^{20} {J}\) . If this energy were delivered to the iceberg every year, how many years would it take before the ice melted?

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, 0.6 kg of blood flows to the body鈥檚 surface and releases 2000 J of energy. The blood arriving at the surface has the temperature of the body鈥檚 interior, \(37.0^{\circ} {C}\) . Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.

When it rains, water vapor in the air condenses into liquid water, and energy is released. (a) How much energy is released when 0.0254 m (one inch) of rain falls over an area of \(2.59 \times 10^{6} {m}^{2}\) (one square mile)? (b) If the average energy needed to heat one home for a year is \(1.50 \times 10^{11} {J}\) , how many homes could be heated for a year with the energy determined in part (a)?

The vapor pressure of water at \(10^{\circ} {C}\) is 1300 \({Pa}\) . (a) What percentage of atmospheric pressure is this? Take atmospheric pressure to be \(1.013 \times 10^{5} {Pa}\) . (b) What percentage of the total air pressure at \(10^{\circ} {C}\) is due to water vapor when the relative humidity is 100\(\% ?\) (c) The vapor pressure of water at \(35^{\circ} {C}\) is 5500 Pa. What is the relative humidity at this temperature if the partial pressure of water in the air has not changed from what it was at \(10^{\circ} {C}\) when the relative humidity was 100\(\% ?\)

mmh When resting, a person has a metabolic rate of about \(3.0 \times 10^{5}\) joules per hour. The person is submerged neck-deep into a tub containing \(1.2 \times 10^{3}\) kg of water at \(21.00^{\circ} {C}\) . If the heat from the person goes only into the water, find the water temperature after half an hour.
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