/*! 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 77 A thirsty nurse cools a \(2.00 \... [FREE SOLUTION] | 91Ó°ÊÓ

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A thirsty nurse cools a \(2.00 \mathrm{~L}\) bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass \(0.257 \mathrm{~kg}\) and adding \(0.120 \mathrm{~kg}\) of ice initially at \(-15.0^{\circ} \mathrm{C}\). If the soft drink and mug are initially at \(20.0^{\circ} \mathrm{C},\) what is the final temperature of the system. assuming no heat losses?

Short Answer

Expert verified
The final temperature is approximately 1.2°C.

Step by step solution

01

Identify System Components and Their Initial States

The system consists of three main components: the soft drink, the aluminum mug, and the ice. The initial temperature of the soft drink and the mug is 20.0°C, while the ice is at -15.0°C. We need to calculate the final equilibrium temperature when these components reach thermal equilibrium.
02

Determine Heat Capacity and Related Constants

- The specific heat capacity of water (soft drink) is approximately 4.18 J/g°C. - The specific heat capacity of aluminum (the mug) is 0.897 J/g°C. - The specific heat capacity of ice is 2.09 J/g°C. - The latent heat of fusion for ice is 334 J/g.
03

Calculate Energy Required to Warm the Ice to 0°C

The mass of the ice is 120 g, and its initial temperature is -15.0°C. The energy required to warm the ice to 0°C is calculated by:\[ Q_1 = m_{ ext{ice}} \times c_{ ext{ice}} \times \Delta T = 120 \times 2.09 \times (0 - (-15)) = 3768 \text{ J}. \]
04

Calculate Energy Required for Ice to Melt

To melt the ice at 0°C, we use the latent heat of fusion:\[ Q_2 = m_{ ext{ice}} \times L_f = 120 \times 334 = 40080 \text{ J}. \]
05

Calculate Energy Changes for the Soft Drink and the Mug

Calculate the heat loss of the soft drink as it cools from 20°C to the final temperature (T):\[ Q_3 = m_{ ext{drink}} \times c_{ ext{drink}} \times (T - 20) = 2000 \times 4.18 \times (T - 20). \]For the aluminum mug from 20°C to T:\[ Q_4 = m_{ ext{mug}} \times c_{ ext{mug}} \times (T - 20) = 257 \times 0.897 \times (T - 20). \]
06

Equate Total Heat Changes to Zero and Solve for T

The total heat changes, considering no heat loss to the environment, should sum to zero:\[ 3768 + 40080 - 2000 \times 4.18 \times (20 - T) - 257 \times 0.897 \times (20 - T) = 0. \]Solve this equation for the final temperature, T.

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a key concept when dealing with thermal problems. It describes the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. In this exercise, we deal with three different materials: water (the soft drink), aluminum (the mug), and ice. Each has a different specific heat capacity, reflecting how much heat they each require to change temperatures slightly.

Here's a quick breakdown:
  • Water (and similarly, the soft drink) has a specific heat capacity of 4.18 J/g°C.
  • Aluminum, a common metal used for its lightweight properties, has a specific heat capacity of 0.897 J/g°C.
  • Ice, the solid form of water, has a specific heat capacity of 2.09 J/g°C prior to reaching 0°C and melting.
These values illustrate why different substances heat up or cool down at different rates. The higher the specific heat capacity, the more energy is needed for a temperature change, meaning water cools down slower than aluminum given the same conditions.
Latent Heat of Fusion
Latent heat of fusion is the energy needed to change a substance from solid to liquid at a constant temperature, without changing its temperature. For water, this is a crucial value needed to calculate the energy required for ice to melt. In our scenario, the latent heat of fusion for water is 334 J/g.

This means that for every gram of ice, 334 J of energy are needed to convert it to liquid water at 0°C. This heat does not alter the temperature of the ice; instead, it breaks the molecular bonds that keep the ice in a solid state. In calculations, the energy required for this phase change forms a significant component of the total energy exchanges in the system.
Heat Transfer
Heat transfer is the process of energy moving from a warmer object to a cooler one, until they both reach the same temperature, known as thermal equilibrium. In this exercise, heat transfers occur between the warm soft drink and mug and the cold ice.

Understanding heat transfer involves recognizing:
  • Energy flows from higher to lower temperatures.
  • In a closed system with no external heat losses, the heat lost by warmer components equals the heat gained by cooler ones.
Calculating this involves summing the energy changes from each component as they reach a common equilibrium temperature. If there's no leak to the environment, every joule lost by the warmer parts of the system goes to warming the cooler parts.
Thermal Energy Calculation
Thermal energy calculation integrates all key concepts discussed, such as specific heat capacity, latent heat of fusion, and heat transfer. This process involves determining the energy changes in the system as it moves towards thermal equilibrium.

Here's how it works in steps:
  • Calculate the energy required to heat the ice from its initial temperature to its melting point.
  • Determine the energy required to melt the ice using the latent heat of fusion.
  • Compute the energy released by the soft drink and the aluminum mug as they cool down to a new equilibrium temperature.
You then set up an equation where the total energy gained by the ice equals the total energy lost by the soft drink and the mug, ensuring the sum of heat changes in the system is zero. Solving this provides the final equilibrium temperature of the system.

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Most popular questions from this chapter

In very cold weather, a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what is the amount of heat needed to warm to internal body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the \(0.50 \mathrm{~L}\) of air exchanged with each breath? Assume that the specific heat of air is \(1020 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K})\) and that \(1.0 \mathrm{~L}\) of air has a mass of \(1.3 \mathrm{~g}\). (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

A metal rod is \(40.125 \mathrm{~cm}\) long at \(20.0^{\circ} \mathrm{C}\) and \(40.148 \mathrm{~cm}\) long at \(45.0^{\circ} \mathrm{C}\). Calculate the average coefficient of linear expansion of the rod's material for this temperature range.

A study published in July 2004 indicated that temperature increases in urban areas in the eastern United States are causing plants to bud up to 7 days early compared with plants in rural areas just a few miles away, thereby disrupting biological cycles. Average temperatures in the urban areas were up to \(3.5 \mathrm{C}^{\circ}\) higher than in the rural areas. By what percent will the radiated heat per square meter increase due to such a temperature difference if the rural temperature is \(0^{\circ} \mathrm{C}\) on the average?

Global warming. As the earth warms, sea level will rise due to melting of the polar ice and thermal expansion of the oceans. Estimates of the expected temperature increase vary, but \(3.5 \mathrm{C}^{\circ}\) by the end of the century has been plausibly suggested. If we assume that the temperature of the oceans also increases by this amount, how much will sea level rise by the year 2100 due only to the thermal expansion of the water? Assume, reasonably, that the ocean basins do not expand appreciably. The average depth of the ocean is \(4000 \mathrm{~m}\), and the coefficient of volume expansion of water at \(20^{\circ} \mathrm{C}\) is \(0.207 \times 10^{-3}\left(\mathrm{C}^{\circ}\right)^{-1}\)

You have no doubt noticed that you usually shiver when you get out of the shower. Shivering is the body's way of generating heat to restore its internal temperature to the normal \(37^{\circ} \mathrm{C}\). and it produces approximately \(290 \mathrm{~W}\) of heat power per square meter of body area. A \(68 \mathrm{~kg}(150 \mathrm{lb}), 1.78 \mathrm{~m}(5 \mathrm{ft}, 10\) in.) person has approximately \(1.8 \mathrm{~m}^{2}\) of surface area. How long would this person have to shiver to raise his or her body temperature by \(1.0 \mathrm{C}^{\circ},\) assuming that none of this heat is lost by the body? The specific heat of the body is about \(3500 \mathrm{~J} /(\mathrm{kg} \cdot \mathrm{K}).\)

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