/*! 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 51 A 0.35-kg coffee mug is made fro... [FREE SOLUTION] | 91Ó°ÊÓ

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A 0.35-kg coffee mug is made from a material that has a specific heat capacity of \(920 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\) and contains \(0.25 \mathrm{kg}\) of water. The cup and water are at \(15^{\circ} \mathrm{C} .\) To make a cup of coffee, a small electric heater is immersed in the water and brings it to a boil in three minutes. Assume that the cup and water always have the same temperature and determine the minimum power rating of this heater.

Short Answer

Expert verified
The minimum power rating of the heater is 646.46 W.

Step by step solution

01

Define the Heat Transfer Formula

To find the power required by the heater, we first need to calculate the energy needed to raise the temperature of the coffee mug and the water from \(15^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) (boiling point of water). This energy can be calculated using the formula: \(Q = mc\Delta T\), where \(Q\) is the heat energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change. We will apply this formula separately for the mug and the water.
02

Calculate Energy for Water

For the water, the mass \(m = 0.25 \mathrm{kg}\) and specific heat capacity \(c = 4186 \mathrm{J} / (\mathrm{kg} \cdot \mathrm{C}^{\circ})\). The temperature change \(\Delta T = 100^{\circ} \mathrm{C} - 15^{\circ} \mathrm{C} = 85^{\circ} \mathrm{C}\). Thus, the energy required is \(Q_{\text{water}} = 0.25 \times 4186 \times 85 = 88992.5 \mathrm{J}\).
03

Calculate Energy for Coffee Mug

For the coffee mug, mass \(m = 0.35 \mathrm{kg}\) and specific heat capacity \(c = 920 \mathrm{J} / (\mathrm{kg} \cdot \mathrm{C}^{\circ})\). The temperature change is also \(85^{\circ} \mathrm{C}\). The energy required is \(Q_{\text{mug}} = 0.35 \times 920 \times 85 = 27370 \mathrm{J}\).
04

Total Energy Required by Heater

The total energy required to heat both the water and the mug is the sum of the energies calculated: \(Q_{\text{total}} = Q_{\text{water}} + Q_{\text{mug}} = 88992.5 + 27370 = 116362.5 \mathrm{J}\).
05

Calculate Power of the Heater

Power is the rate of energy transfer, calculated using the formula \(P = \frac{Q}{t}\), where \(t\) is the time in seconds. Given that the heater heats the mug and water in 3 minutes (or 180 seconds), the power needed is \(P = \frac{116362.5}{180} = 646.46 \mathrm{W}\).

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Key Concepts

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

Understanding Specific Heat Capacity
Specific heat capacity is a fundamental concept in thermodynamics that describes how much heat energy a substance can absorb before it changes temperature. It's like a measure of how "heat retentive" a material is. Each material has its specific heat capacity, denoted by the symbol \( c \), and this value tells us how much energy (in joules) is required to raise the temperature of one kilogram of the substance by one degree Celsius.
In our exercise, the coffee mug and water both require heat to increase their temperatures from \( 15^{\circ} \mathrm{C} \) to \( 100^{\circ} \mathrm{C} \). For the mug, its specific heat capacity is \( 920 \mathrm{J} / (\mathrm{kg} \cdot \mathrm{C}^{\circ}) \). This value is significant in calculating the amount of energy needed for heating. The higher the specific heat capacity, the more energy you'll need to change the temperature.
  • High Capacity: Substances like water, which have a high specific heat capacity (\( 4186 \mathrm{J} / (\mathrm{kg} \cdot \mathrm{C}^{\circ}) \)), absorb more energy without a big temperature change. This makes them excellent for storing and transporting heat.
  • Low Capacity: Conversely, metals typically have low specific heat capacities, meaning they heat up and cool down quickly.
The specific heat capacity plays a key role in determining how different substances will react to heat energy — ensuring accurate thermal management in practical applications.
Power Calculation Basics
Calculating power is about understanding the rate at which energy is used. In physics, power is defined as the amount of energy transferred or converted per unit of time. The formula for power is \( P = \frac{Q}{t} \), where \( P \) is the power in watts, \( Q \) is the energy in joules, and \( t \) is the time in seconds.
Using the exercise, we need to know how much power the heater must have to bring the coffee mug and water to a boil in three minutes (or 180 seconds). We already calculated the total heat energy needed, \( Q_{\text{total}} = 116362.5 \mathrm{J} \). Now, we're figuring out how quickly that energy must be supplied:
  • Power Calculation: Substitute the values into the formula: \( P = \frac{116362.5}{180} \).
  • Result: The calculated power is \( 646.46 \mathrm{W} \). This means the heater needs a minimum power rating of \( 646.46 \mathrm{W} \) to efficiently perform the task within the given time.
Understanding power calculations helps us design systems that effectively balance energy efficiency and performance — crucial for everything from household appliances to industrial machinery.
Temperature Change and Its Effects
Temperature change is an essential concept that shows how the temperature of a substance varies when heat energy is applied. In thermodynamics, this is indicated by \( \Delta T \), which is the difference between the final and initial temperatures of a substance.
In the coffee mug scenario, we're interested in changing the temperature from \( 15^{\circ} \mathrm{C} \) to the boiling point of water, \( 100^{\circ} \mathrm{C} \), resulting in a temperature change of \( 85^{\circ} \mathrm{C} \). This change helps us understand how much energy is needed to achieve this transformation.
Let's break down why it's crucial:
  • Energy Requirement: The greater the temperature change, the greater the amount of energy required. This is directly used in the formula \( Q = mc\Delta T \), where \( \Delta T \) significantly impacts the heat needed.
  • Practical Implications: Knowing \( \Delta T \) allows us to plan heating or cooling processes effectively. For instance, we need more energy to achieve larger temperature changes in heating applications.
Temperature change isn't just a number; it's a key parameter influencing how we manage energy resources in both daily life and advanced technological settings.

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Most popular questions from this chapter

A \(0.200-\mathrm{kg}\) piece of aluminum that has a temperature of \(-155^{\circ} \mathrm{C}\) is added to \(1.5 \mathrm{kg}\) of water that has a temperature of \(3.0^{\circ} \mathrm{C}\). At equilibrium the temperature is \(0.0^{\circ} \mathrm{C}\). Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

When it rains, water vapor in the air condenses into liquid water, and energy is released. (a) How much energy is released when \(0.0254 \mathrm{m}\) (one inch) of rain falls over an area of \(2.59 \times 10^{6} \mathrm{m}^{2}\) (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)?

What's your normal body temperature? It may not be \(98.6^{\circ} \mathrm{F}\), the often-quoted average that was determined in the nineteenth century. A more recent study has reported an average temperature of \(98.2^{\circ} \mathrm{F}\). What is the difference between these averages, expressed in Celsius degrees?

The latent heat of vaporization of \(\mathrm{H}_{2} \mathrm{O}\) at body temperature \(\left(37.0^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} .\) To cool the body of a \(75-\mathrm{kg}\) jogger [average specific heat capacity \(\left.=3500 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\right]\) by \(1.5 \mathrm{C}^{\circ},\) how many kilograms of water in the form of sweat have to be evaporated?

The heating element of a water heater in an apartment building has a maximum power output of \(28 \mathrm{kW}\). Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\). If the water going into the heater has a temperature of \(11{ }^{\circ} \mathrm{C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

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