/*! 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 24 In an effort to stay awake for a... [FREE SOLUTION] | 91Ó°ÊÓ

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In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a \(200 \mathrm{~W}\) electric immersion heater in \(0.320 \mathrm{~kg}\) of water. (a) How much heat must be added to the water to raise its temperature from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C} ?\) (b) How much time is required? Assume that all of the heater's power goes into heating the water.

Short Answer

Expert verified
The heat required to raise the water's temperature is 80.0 kJ and it would take the heater approximately 6.7 minutes.

Step by step solution

01

Calculate the heat required to raise the temperature

Firstly, we need to calculate the heat required to raise the temperature of water from \(20.0^{\circ} \mathrm{C}\) to \(80.0^{\circ} \mathrm{C}\). This is done using the heat formula \(q=mc \Delta T\), where \(m = 0.320 \mathrm{~kg}\) (the mass of the water), \(c = 4.186 \mathrm{~J/g \cdot C}\) (specific heat of water) and \(\Delta T = 80.0 - 20.0 = 60 \mathrm{~C}\) (the change in temperature). Note that we need to convert the mass from kg to g.
02

Calculate the time required

To define the time it would take for the heater to raise the water's temperature, we can use the formula for power, \(P = \frac{q}{t}\). We know that all the heater's power (\(200 \mathrm{~W}\) or \(200 \mathrm{~Js^{-1}}\)) goes into heating the water, and we have calculated the required heat in Step 1. Thus, we can rearrange the formula to solve for time, \(t = \frac{q}{P}\).
03

Convert time to minutes

The time calculated in Step 2 will be in seconds. To convert it to minutes, simply divide the time in seconds by 60.

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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 measure of how much energy it takes to raise the temperature of a substance by a certain amount. It is a property that is unique to each substance and is usually denoted by the symbol 'c'. In our exercise, the specific heat capacity is used to calculate the amount of energy needed to heat water from one temperature to another.

For an intimate understanding, consider specific heat capacity as a personality trait of a material. Just like some people are more sensitive to cold than others, different materials need more or less energy to change their temperatures. Water, for example, has a high specific heat capacity, meaning it requires a significant amount of energy to increase its temperature. This is why water is great for cooling systems or why it takes longer to boil a pot of water compared to getting a metal pan hot.

Mathematically, we calculate the heat (q) absorbed or released using the formula: \(q = mc\Delta T\). Here, 'm' stands for the mass of the substance, 'c' is the specific heat capacity, and \(\Delta T\) is the change in temperature (final temperature minus initial temperature). For our sleepy student, knowing how much energy they have to spend to heat their water is crucial for determining how long their study night will last!
Heat Formula
The heat formula is essential for solving problems involving thermal energy transfer. It's the equation \(q = mc\Delta T\) that we've introduced above. This formula helps to determine the amount of energy, in joules, required to change the temperature of a certain mass of a substance by a particular amount.

Breaking down the heat formula piece by piece makes it easier to understand. The 'q' represents the heat energy that is either absorbed or released. The 'm' is the mass of the substance in question, and it's crucial because the more of a substance you have, the more energy it will need to change its temperature. The 'c' is the specific heat capacity, as discussed earlier. It's like a conversion factor that tells us how many joules of energy are required to raise one gram of the substance by one degree Celsius. Finally, \(\Delta T\) is the temperature change – simply the difference between where you're starting and where you're ending.

This formula is immensely useful, not just in the study room for an upcoming exam but also in industries where heating or cooling processes are essential. By rearranging the heat formula, engineers can determine how much energy they need to achieve the desired temperature change in a system.
Electric Immersion Heaters
Electric immersion heaters are devices designed to heat liquids directly through electrical energy. They are typically used to heat water in tanks, swimming pools, and, as seen in our example, for a student's cup of coffee during an all-nighter. The immersion heater works by converting electrical energy into heat, which is then transferred to the surrounding liquid.

An immersion heater's efficiency is maximum when all of its energy goes into heating the intended substance, with minimal losses to the surrounding environment. In the case of our diligent student, it's noted that the heater's entire power output – which is 200 watts, or 200 joules per second – is being used to heat the water. This assumption simplifies the calculations as there are no energy losses to consider.

How Immersion Heaters Work

When you turn on an immersion heater, electricity passes through a resistive heating element. Resistance causes the element to generate heat, which is then transferred to the water. The basic premise here is similar to how a toaster works to brown your bread, but instead of crisping bread, we're cozying up water molecules. By using the specific heat capacity of water and the heat formula in conjunction with the heater's power, one can precisely calculate the time needed to reach the desired temperature. It's a brilliant practical application of physics in everyday life, ensuring our student remains alert for the academic challenges that await!

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Most popular questions from this chapter

Convert the following Celsius temperatures to Fahrenheit: (a) \(-62.8^{\circ} \mathrm{C}\), the lowest temperature ever recorded in North America (February \(3,1947,\) Snag, Yukon); (b) \(56.7^{\circ} \mathrm{C},\) the highest temperature ever recorded in the United States (July \(10,1913,\) Death Valley, California); (c) \(31.1^{\circ} \mathrm{C},\) the world's highest average annual temperature (Lugh Ferrandi, Somalia).

Evaporation of sweat is an important mechanism for temperature control in some warm-blooded animals. (a) What mass of water must evaporate from the skin of a \(70.0 \mathrm{~kg}\) man to cool his body \(1.00 \mathrm{C}^{\circ}\) ? The heat of vaporization of water at body temperature \(\left(37^{\circ} \mathrm{C}\right)\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) The specific heat of a typical human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) (see Exercise 17.27 ) . (b) What volume of water must the man drink to replenish the evaporated water? Compare to the volume of a soft-drink can \(\left(355 \mathrm{~cm}^{3}\right)\).

While running, a \(70 \mathrm{~kg}\) student generates thermal energy at a rate of \(1200 \mathrm{~W}\). For the runner to maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the energy could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K},\) slightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

What is the rate of energy radiation per unit area of a blackbody at (a) \(273 \mathrm{~K}\) and (b) \(2730 \mathrm{~K} ?\)

\(17.76 \bullet\) A surveyor's \(30.0 \mathrm{~m}\) steel tape is correct at \(20.0^{\circ} \mathrm{C}\). The distance between two points, as measured by this tape on a day when its temperature is \(5.00^{\circ} \mathrm{C},\) is \(25.970 \mathrm{~m} .\) What is the true distance between the points?

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