/*! 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

91Ó°ÊÓ

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.

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.

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!

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

A glass flask whose volume is \(1000.00 \mathrm{~cm}^{3}\) at \(0.0^{\circ} \mathrm{C}\) is completely filled with mercury at this temperature. When flask and mercury are warmed to \(55.0^{\circ} \mathrm{C}, 8.95 \mathrm{~cm}^{3}\) of mercury overflow. If the coefficient of volume expansion of mercury is \(18.0 \times 10^{-5} \mathrm{~K}^{-1},\) compute the coefficient of volume expansion of the glass.

One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\). To treat a \(70.0 \mathrm{~kg}\) patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C}\).

A \(\mathrm 500.0 \mathrm{~g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing \(1.00 \mathrm{~kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\). (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

If the air temperature is the same as the temperature of your skin (about \(30^{\circ} \mathrm{C}\) ), your body cannot get rid of heat by transferring it to the air. In that case, it gets rid of the heat by evaporating water (sweat). During bicycling, a typical \(70 \mathrm{~kg}\) person's body produces energy at a rate of about \(500 \mathrm{~W}\) due to metabolism, \(80 \%\) of which is converted to heat. (a) How many kilograms of water must the person's body evaporate in an hour to get rid of this heat? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (b) The evaporated water must, of course, be replenished, or the person will dehydrate. How many \(750 \mathrm{~mL}\) bottles of water must the bicyclist drink per hour to replenish the lost water? (Recall that the mass of a liter of water is \(1.0 \mathrm{~kg} .\) )

The emissivity of tungsten is 0.350 . A tungsten sphere with radius \(1.50 \mathrm{~cm}\) is suspended within a large evacuated enclosure whose walls are at \(290.0 \mathrm{~K}\). What power input is required to maintain the sphere at \(3000.0 \mathrm{~K}\) if heat conduction along the supports is ignored?

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.