/*! 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 121 Thermal energy is added to \(180... [FREE SOLUTION] | 91Ó°ÊÓ

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Thermal energy is added to \(180 \mathrm{~g}\) of water at a constant rate for \(3.5 \mathrm{~min}\), resulting in an increase in temperature of \(12^{\circ} \mathrm{C}\). What is the heating rate, in joules per second?

Short Answer

Expert verified
The heating rate is approximately \(51.84 \text{ J/s}\).

Step by step solution

01

Identify Given Values

We are given the following values: mass of water, \(m = 180 \text{ g}\); time, \(t = 3.5 \text{ min}\); temperature increase, \(\Delta T = 12^{\circ} \text{C}\). We need to find the heating rate in joules per second.
02

Convert Units

Convert the mass from grams to kilograms as \(m = 0.18 \text{ kg}\), and time from minutes to seconds as \(t = 3.5 \times 60 = 210 \text{ seconds}\).
03

Use Specific Heat Formula

Use the formula for heat transferred \(Q = mc\Delta T\), where \(c\) is the specific heat capacity of water, \(c = 4200 \text{ J/kg}\text{°C}\).
04

Calculate Heat Transferred

Substitute the known values into the formula: \(Q = 0.18 \times 4200 \times 12\). Calculate \(Q = 907.2 \times 12 = 10886.4 \text{ Joules}\).
05

Calculate Heating Rate

The heating rate is \(\frac{Q}{t} = \frac{10886.4}{210}\). Calculate \( \approx 51.84 \text{ J/s} \).

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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 understanding how substances respond to heat. It tells us how much energy is needed to change the temperature of a unit mass of a substance by one degree Celsius.
For water, this value is quite high, at 4200 J/kg°C. This means water can absorb a lot of heat before it even begins to change temperature. This property makes water an excellent substance for regulating temperatures, ideal for cooling systems and weather moderation.
When dealing with problems involving heat transfer, you'll often see the formula:
  • \( Q = mc\Delta T \)
Here, \(Q\) is the thermal energy transferred, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.
Knowing specific heat capacity allows us to calculate how much thermal energy is required to achieve a desired temperature change, a crucial insight for a wide range of practical applications.
Exploring Thermal Energy
Thermal energy is the energy that comes from the temperature of matter. It is generated from the movement of particles within an object. The faster these particles move, the more thermal energy the object has.
In our exercise, we explored how much thermal energy was required to raise the temperature of water. The calculation involved was based on the specific heat capacity formula, which helped us decide how much energy was transferred to the water when it was heated over time.
Thermal energy includes stored energy in chemical bonds and kinetic energy from moving particles. Here, converting this energy into heat causes a discernible change in temperature, translating into tangible effects like heating water or cooking food. Understanding how thermal energy works helps us grasp the complexity of other energy transformations and utilize them effectively in daily life.
Grasping Temperature Change
Temperature change serves as an indicator of heat transfer into or out of a system. A positive temperature change means a system absorbs heat; a negative change indicates heat loss.
In the exercise, the 12°C increase signified that water absorbed thermal energy. Such knowledge is pivotal for calculating heating and cooling requirements in various scenarios.
The
  • equation \(\Delta T = \frac{Q}{mc} \)
can determine temperature changes from known quantities of heat added or removed. Understanding this lets us anticipate how substances will respond to energy inputs and guides critical decisions in engineering, meteorology, and culinary arts, among others. These practical applications illustrate why mastering temperature change is essential for efficient energy use.

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Most popular questions from this chapter

To determine the specific heat capacity of an object, a student heats it to \(100^{\circ} \mathrm{C}\) in boiling water. She then places the \(38.0\)-g object in a 155 -g aluminum calorimeter containing \(103 \mathrm{~g}\) of water. The aluminum and water are initially at a temperature of \(20.0^{\circ} \mathrm{C}\) and are thermally insulated from their surroundings. If the final temperature is \(22.0^{\circ} \mathrm{C}\), what is the specific heat capacity of the object? Referring to Table \(10.2\), identify the material that the object is made of.

A system consists of a \(0.130-\mathrm{kg}\) chunk of ice floating in \(1.12 \mathrm{~kg}\) of water, all at \(0{ }^{\circ} \mathrm{C}\). How much thermal energy must be added to the system to convert it to \(1.25 \mathrm{~kg}\) of water at \(15.0^{\circ} \mathrm{C}\) ?

Follow-up Find the Celsius temperature that corresponds to \(110^{\circ} \mathrm{F}\).

Predict \& Explain The specific heat capacity of isopropyl alcohol is about half that of water. Suppose you have \(0.5 \mathrm{~kg}\) of isopropyl alcohol at \(20^{\circ} \mathrm{C}\) in one container and \(0.5 \mathrm{~kg}\) of water at \(30^{\circ} \mathrm{C}\) in a second container. (a) When these fluids are poured into the same container and allowed to come to thermal equilibrium, is the final temperature greater than, less than, or equal to \(25^{\circ} \mathrm{C}\) ? (b) Choose the best explanation from among the following: A. The low specific heat capacity of isopropyl alcohol means that it accepts more thermal energy, giving a final temperature that is less than \(25^{\circ} \mathrm{C}\). B. More thermal energy is required to change the temperature of water than to change the temperature of isopropyl alcohol. Therefore, the final temperature will be greater than \(25^{\circ} \mathrm{C}\). C. Equal masses are mixed together; therefore, the final temperature will be \(25^{\circ} \mathrm{C}\), the average of the two initial temperatures.

Calculate A metal rod has an initial length of \(2.5 \mathrm{~m}\). When its temperature is increased by \(85^{\circ} \mathrm{C}\), its length increases by \(0.36 \mathrm{~cm}\). What is the metal's coefficient of thermal expansion? Which metal is it most likely to be?

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