/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 (II) High-altitude mountain clim... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(II) High-altitude mountain climbers do not eat snow, but always melt it first with a stove. To see why, calculate the energy absorbed from your body if you \((a)\) eat \(1.0 \mathrm{~kg}\) of \(-10^{\circ} \mathrm{C}\) snow which your body warms to body temperature of \(37^{\circ} \mathrm{C}\). ( \(b\) ) You melt \(1.0 \mathrm{~kg}\) of \(-10^{\circ} \mathrm{C}\) snow using a stove and drink the resulting \(1.0 \mathrm{~kg}\) of water at \(2^{\circ} \mathrm{C}\), which your body has to warm to \(37^{\circ} \mathrm{C}\)

Short Answer

Expert verified
Eating snow directly requires 509.66 kJ, but melting and drinking it as water only requires 146.3 kJ. Melting is more energy efficient.

Step by step solution

01

Calculating energy to warm snow from -10°C to 0°C

First, we need to calculate the energy required to warm the snow from -10°C to 0°C. The formula to calculate the heat absorbed is given by \( Q = mc\Delta T \). Here, \( m = 1.0 \text{ kg} \), \( c = 2.1 \text{ kJ/kg°C} \) (specific heat capacity of ice), and \( \Delta T = 0°C - (-10°C) = 10°C \). Therefore, \( Q = 1.0 \times 2.1 \times 10 = 21 \text{ kJ} \).
02

Calculating energy to melt the snow

Next, we calculate the energy required to melt the snow at 0°C into water. The latent heat of fusion of ice is \( 334 \text{ kJ/kg} \). Thus, the energy required is \( Q = mL = 1.0 \times 334 = 334 \text{ kJ} \).
03

Calculating energy to warm water from 0°C to 37°C

Now we calculate the energy required to warm the water from 0°C to 37°C. Using \( Q = mc\Delta T \), where \( c = 4.18 \text{ kJ/kg°C} \) (specific heat capacity of water), the temperature change \( \Delta T = 37°C - 0°C = 37°C \). The energy needed is \( Q = 1.0 \times 4.18 \times 37 = 154.66 \text{ kJ} \).
04

Total energy absorbed if eating snow

Add all the energies from Steps 1-3 to find the total energy absorbed by the body when eating the snow directly. Total \( Q = 21 + 334 + 154.66 = 509.66 \text{ kJ} \).
05

Calculating energy to warm water from 2°C to 37°C

If the snow is melted and drank as water at 2°C, calculate the energy required to warm from 2°C to 37°C. Using \( Q = mc\Delta T \), where \( \Delta T = 37°C - 2°C = 35°C \), the energy needed is \( Q = 1.0 \times 4.18 \times 35 = 146.3 \text{ kJ} \).
06

Comparing both methods

Compare the energy required in each method. Eating snow directly requires 509.66 kJ, whereas melting it first and then consuming only requires 146.3 kJ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Heat Capacity
When studying thermodynamics, it's essential to understand heat capacity. Heat capacity is a property that describes the amount of heat energy required to change a substance's temperature by a certain amount. It is detailed by the formula \( Q = mc\Delta T \), where \( Q \) is the heat added, \( m \) the mass, \( c \) the specific heat capacity, and \( \Delta T \) the change in temperature. This is crucial for processes like warming snow from \(-10°C\) to \(0°C\), where the specific heat capacity of ice \( c \) is \(2.1 \text{ kJ/kg°C}\).
The concept of heat capacity helps us calculate how much energy the body has to expend when consuming cold substances. In our example, the body has to put out 21 kJ of energy just to bring the temperature of the snow up from \(-10°C\) to \(0°C\). This illustrates why mountain climbers prefer pre-melting snow; their bodies can conserve energy for more vital functions. This principle applies not only to melting snow but also to heating any other materials within thermodynamic systems.
Latent Heat
Latent heat is another critical concept in thermodynamics. It refers to the amount of heat required to change the state of a given amount of substance without changing its temperature. For ice turning into water, this is called the latent heat of fusion. The amount of heat required is described by \( Q = mL \), where \( L \) is the latent heat of fusion.
In our exercise, once the snow (ice) reaches \(0°C\), it needs extra energy to change into water while remaining at the same temperature. This energy, in this case, totals 334 kJ for 1 kg of ice. This phase transition requires substantial energy, explaining why it makes sense to use a stove to melt the snow instead of relying on body heat alone. Understanding latent heat is critical when evaluating energy transformations in any physical or chemical process involving phase changes.
Energy Transfer
Energy transfer plays a pivotal role in thermodynamics and is the process through which energy is transferred from one system to another, typically as heat. Understanding energy transfer helps us determine how much energy a body or system needs to adjust or maintain a desired state or condition.
In our context, whether mountain climbers eat snow or drink melted water, energy transfer occurs to bring the substance to the body's core temperature of \(37°C\). Eating snow requires conducting three significant energy transformations: heating the ice to its melting point, melting it, and then heating the water. When snow is melted first, only one transformation needs to occur, warming the water from \(2°C\) to \(37°C\). This process requires only 146.3 kJ, significantly less energy than the combined transformations of 509.66 kJ required if the snow were consumed directly. This highlights how manipulating energy transfer processes can lead to more efficient and less energy-intensive solutions in various scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(III) A 1.00 -mol sample of an ideal diatomic gas at a pressure of 1.00 atm and temperature of \(420 \mathrm{~K}\) undergoes a process in which its pressure increases linearly with temperature. The final temperature and pressure are \(720 \mathrm{~K}\) and 1.60 atm. Determine \((a)\) the change in internal energy, (b) the work done by the gas, and (c) the heat added to the gas. (Assume five active degrees of freedom.)

(I) An automobile cooling system holds 18 L of water. How much heat does it absorb if its temperature rises from \(15^{\circ} \mathrm{C}\) to \(95^{\circ} \mathrm{C}\) ?

(II) In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, \(2850 \mathrm{~J}\) of work is done on the gas. ( \(a\) ) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall?

(III) In the process of taking a gas from state a to state \(\mathrm{c}\) along the curved path shown in Fig. \(19-32,85 \mathrm{~J}\) of heat leaves the system and \(55 \mathrm{~J}\) of work is done on the system. (a) Determine the change in internal energy, \(E_{\text {int, a }}-E_{\text {int, c }}\) (b) When the gas is taken along the path cda, the work done by the gas is \(W=38 \mathrm{~J}\). How much heat \(Q\) is added to the gas in the process cda? (c) If \(P_{\mathrm{a}}=2.2 P_{\mathrm{d}},\) how much work is done by the gas in the process abc? ( \(d\) ) What is \(Q\) for path abc? \((e)\) If \(E_{\text {int, a }}-E_{\text {int }, \mathrm{b}}=15 \mathrm{~J},\) what is \(Q\) for the process bc? Here is a summary of what is given: $$ \begin{aligned} Q_{\mathrm{a} \rightarrow \mathrm{c}} &=-85 \mathrm{~J} \\ W_{\mathrm{a}->\mathrm{c}} &=-55 \mathrm{~J} \\ W_{\mathrm{cda}} &=38 \mathrm{~J} \\ E_{\text {int, } \mathrm{a}}-E_{\text {int }, \mathrm{b}} &=15 \mathrm{~J} \\ P_{\mathrm{a}} &=2.2 P_{\mathrm{d}} . \end{aligned} $$

(I) \((a)\) How much power is radiated by a tungsten sphere (emissivity \(\epsilon=0.35\) of radius 16 \(\mathrm{cm}\) at a temperature of \(25^{\circ} \mathrm{C} ?(b)\) If the sphere is enclosed in a room whose walls are kept at \(-5^{\circ} \mathrm{C},\) what is the net flow rate of energy out of the sphere?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.