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Chapter 2: Problem 153

Dieters are often told that drinking ice-cold water burns more energy than drinking roomtemperature water. Why is this true?

Short Answer

Expert verified
Drinking ice-cold water burns more energy than drinking room temperature water because our body needs to transfer more energy (in the form of heat) to the cold water to reach thermal equilibrium. This process of transferring heat from the body to the water requires burning more calories, which is why dieters are often advised to consume ice-cold water.

Step by step solution

01

Understanding energy and temperature

Temperature is a measure of the average kinetic energy of the particles in a substance. When two substances with different temperatures come into contact, energy in the form of heat transfers from the substance with a higher temperature to the one with lower temperature. This process continues until both substances reach thermal equilibrium, meaning they have the same temperature.
02

Drinking ice-cold water

When someone drinks ice-cold water, the water has a lower temperature than the body. Thus, the water absorbs heat from the body to reach thermal equilibrium. This process of the body transferring heat to the water requires energy. Our body gets energy from the food we eat, and this energy is often measured in calories. So, burning calories in this context means using up energy to warm up the ice-cold water inside the body.
03

Drinking room temperature water

When drinking room temperature water, there may still be some heat transfer happening, but as the water is closer in temperature to the body, it requires less energy to bring it to equilibrium. This means that fewer calories will be burned compared to when drinking ice-cold water.
04

Conclusion

Drinking ice-cold water burns more energy than drinking room temperature water because our body needs to transfer more energy (in the form of heat) to the cold water to reach thermal equilibrium. This process of transferring heat from the body to the water requires burning more calories, which is why dieters are often advised to consume ice-cold water.

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Most popular questions from this chapter

A metal sphere has a radius \(r\) of \(4.00 \mathrm{~cm}\). What is the volume \(V\) of this sphere in cubic centimeters? The formula for the volume of a sphere is \(V=(4 / 3) \pi r^{3}\), where \(\pi=3.14159 .\)

Gold has a density of \(19.3 \mathrm{~g} / \mathrm{mL}\). Suppose you have \(100.0\) glonkins of gold. What volume in liters will the gold occupy? Here are some conversion factors to help you: \(0.911\) ounce per glonkin and \(28.35 \mathrm{~g}\) per ounce. Use unit analysis to calculate your answer, and show your work. Treat both conversion factors as exact.

Convert: (a) \(7.98 \times 10^{23} \mu \mathrm{L}\) to liters (b) \(3.00 \times 10^{-3} \mathrm{mg}\) to grams (c) \(4.21 \times 10^{8} \mathrm{~mL}\) to gallons \(\left[1 \mathrm{~m}^{3}=264\right.\) gallons \(]\)

Convert: (a) \(4.50\) Cal to calories (b) \(600.0\) Cal to kilojoules (c) \(1.000 \mathrm{~J}\) to calories (d) \(50.0\) Cal to joules

Do the following calculations and express each answer to the correct number of significant figures. (All values are measurements.) (a) \(\frac{5.03+7.2}{0.003}\) (b) \(\frac{8.93 \times 0.054}{1.32}\) (c) \((6.23 \times 0.042)+9.86\)
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