Question
A physics student foolishly wants to lose weight by drinking cold water. If he drinks $1 \mathrm{L}\left(1000 \mathrm{cm}^{3}\right)$ of water at $10^{\circ} \mathrm{C}$ below body temperature, how many Calories will it take to warm the water?
Step 1
We know that 1 liter of water is equivalent to 1 kilogram. So, the mass of the water is 1 kilogram. \[ m = 1 \, \text{kg} \] Show more…
Show all steps
Your feedback will help us improve your experience
David Zhang and 92 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
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 “burn calories?” 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.
A certain diet doctor encourages people to diet by drinking ice water. His theory is that the body must burn off enough fat to raise the temperature of the water from $0.00^{\circ} \mathrm{C}$ to the body temperature of $37.0^{\circ} \mathrm{C}$ How many liters of ice water would have to be consumed to burn off 454 $\mathrm{g}$ (about 1 1 $\mathrm{b} )$ of fat, assuming that burning this much fat requires 3500 Cal be transferred to the ice water? Why is it not advisable to follow this diet? (One liter $=10^{3} \mathrm{cm}^{3}$ . The density of water is 1.00 $\mathrm{g} / \mathrm{cm}^{3} . )$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD