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Chapter 17: Problem 26

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 is 80,371.2 J, and the time required is about 402 seconds.

Step by step solution

01

Identify the Given Information

We are given that the power of the heater is 200 W, the mass of the water is 0.320 kg, the initial temperature is 20.0 掳C, and the final temperature is 80.0 掳C. The specific heat capacity of water is approximately 4,186 J/(kg路掳C).
02

Calculate the Heat Required

To calculate the heat \(Q\) required, use the formula: \[ Q = m \, c \, \Delta T \]where \(m\) is the mass of the water, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Substitute the given values:\[ Q = 0.320 \, \text{kg} \times 4,186 \, \text{J/(kg路掳C)} \times (80.0 \, 掳C - 20.0 \, 掳C) \]\[ Q = 0.320 \times 4,186 \times 60 \]\[ Q = 80,371.2 \, \text{J} \]
03

Calculate the Time Required

To find the time \(t\) required, use the formula: \[ Q = P \, t \]where \(P\) is the power of the heater. Rearrange the formula to solve for time: \[ t = \frac{Q}{P} \]Substitute the values calculated previously:\[ t = \frac{80,371.2 \, \text{J}}{200 \, \text{W}} \]\[ t = 401.856 \, \text{s} \]Thus, it takes approximately 402 seconds.

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Key Concepts

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

Heat Transfer
Understanding heat transfer is crucial when dealing with temperature changes in thermodynamics. It is the movement of thermal energy from a substance at a higher temperature to one at a lower temperature.
Basically, heat transfer occurs in three main ways: conduction, convection, and radiation. Conduction involves direct transfer through contact, like when the heater warms the water. Convection occurs in fluids, where warmer areas rise and cooler areas sink, creating a circulation pattern. Radiation involves the transfer of energy through electromagnetic waves, like sunlight warming the Earth.
The fundamental principle here is that heat will flow from the hotter material to the cooler one until thermal equilibrium is reached. In our exercise, the heat supplied by the heater is transferred to the water, raising its temperature.
Specific Heat Capacity
The concept of specific heat capacity is essential when calculating how much heat energy is required to change a substance's temperature. It is defined as the amount of heat needed to raise the temperature of 1 kilogram of a substance by 1 degree Celsius.
The formula to calculate heat using specific heat capacity is:
  • \( Q = m \, c \, \Delta T \)
Here,
  • \( Q \) is the heat energy in joules
  • \( m \) is the mass in kilograms
  • \( c \) is the specific heat capacity in joules per kilogram per degree Celsius
  • \( \Delta T \) is the change in temperature
For water, which has a specific heat capacity of approximately 4,186 J/(kg路掳C), the calculations show the importance of this property in heating applications. It signifies that water can absorb a lot of heat before its temperature rises significantly, making it an effective coolant.
Energy Conversion
Energy conversion is a fundamental concept in thermodynamics, describing how energy changes from one form to another. In many practical situations, mechanical, electrical, or chemical energy is converted into thermal energy, as in the example of the electric immersion heater.
This conversion process is vital in everyday appliances, where electrical energy is consumed to produce heat for cooking, heating, or industrial processes. The formula
  • \( Q = P \, t \)
is used to calculate the time required to achieve the desired heat output after energy conversion. Here,
  • \( Q \) represents the heat energy,
  • \( P \) represents the power in watts (the rate of energy conversion),
  • \( t \) represents time in seconds.
In our exercise, the electrical heater converts 200 watts of electrical power into heat energy, effectively warming the water by transferring and converting energy to meet the set thermal criteria.

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Most popular questions from this chapter

Treatment for a Stroke. 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 -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 \(37.0^{\circ} \mathrm{C} .\)

While painting the top of an antenna 225 \(\mathrm{m}\) in height, a worker accidentally lets a \(1.00-\mathrm{L}\) wattle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

Why Do the Seasons Lag? In the northern hemisphere, June 21 (the summer solstice) is both the longest day of the year and the day on which the sun's rays strike the earth most vertically, hence delivering the greatest amount of heat to the surface. Yet the hottest summer weather usually occurs about a month or so later. Let us see why this is the case. Because of the large specific heat of water, the oceans are slower to warm up than the land (and also slower to cool off in winter). In addition to perusing pertinent information in the tables included in this book, it is useful to know that approximately two-thirds of the earth's surface is ocean composed of salt water having a specific heat of 3890 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that the oceans, on the average, are 4000 m deep. Typically, an average of 1050 \(\mathrm{W} / \mathrm{m}^{2}\) of solar energy falls on the earth's surface, and the oceans absorb essentially all of the light that strikes them. However, most of that light is absorbed in the upper 100 \(\mathrm{m}\) of the surface. Depths below that do not change temperature seasonally. Assume that the sunlight falls on the surface for only 12 hours per day and that the ocean retains all the heat it absorbs. What will be the rise in temperature of the upper 100 \(\mathrm{m}\) of the oceans during the month following the summer solstice? Does this seem to be large enough to be perceptible?

Bicycling on a Warm Day. 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 -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 -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 \(\mathrm{kg.}\)

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 \(\mathrm{K}\) . What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and (b) \(\mathrm{C}^{\circ} ?\)
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