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Chapter 11: Problem 11

A \(200-\mathrm{g}\) aluminum cup contains \(800 \mathrm{~g}\) of water in thermal equilibrium with the cup at \(80^{\circ} \mathrm{C}\). The combination of cup and water is cooled uniformly so that the temperature decreases by \(1.5^{\circ} \mathrm{C}\) per minute. At what rate is energy being removed? Express your answer in watts.

Short Answer

Expert verified
The rate at which energy is being removed is approximately -88.4 Watts (the negative sign indicates energy is being removed from the system).

Step by step solution

01

Determine the Specific Heats

Look up or recall the specific heat values for the materials in question, here aluminum and water. The specific heat of aluminum is \(0.897 \mathrm{J / g \cdot ^{\circ}C}\) and for water it is \(4.186 \mathrm{J / g \cdot ^{\circ}C}\).
02

Calculate the Total Heat loss per Minute

The total heat loss due to a decrease in temperature can be calculated using the formula \(\Delta Q = mc\Delta T\) where \(m\) is mass, \(c\) is specific heat and \(\Delta T\) is change in temperature. Since both the cup and the water are losing heat, calculate this for both and sum it up: \(\Delta Q_{total} = (m_{water}c_{water} + m_{al}c_{al})\Delta T\)
03

Plug in Given Values

Insert the given values into the equation from Step 2: \(\Delta Q_{total} = ((800 \, \mathrm{g})(4.186 \, \mathrm{J / g \cdot ^{\circ}C}) + (200 \, \mathrm{g})(0.897 \, \mathrm{J / g \cdot ^{\circ}C}))(1.5 \, ^{\circ}\mathrm{C})\)
04

Simplify

The above calculation will provide the heat loss per minute. Remember that the heat loss is negative since the system is losing heat. After completion, it comes out to \(-5302.7 \, \mathrm{J/min}\)
05

Convert to Watts

1 watt is defined as 1 joule/second, so to convert the result to watts, use the conversion factor 1 minute = 60 seconds: \(-5302.7 \, \mathrm{J/min} \times (1 \, \mathrm{min}/60 \, \mathrm{s}) = -88.38 \, \mathrm{W}\)

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Key Concepts

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

Specific Heat
Understanding specific heat is crucial when studying the rate of heat transfer in materials. Specific heat is a property that tells us how much heat energy a material can absorb before its temperature changes. In scientific terms, it is defined as the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius.

The unit for specific heat is usually in joules per gram per degree Celsius ( ext{J/g} ullet ext{{掳C}}). Different materials have different specific heats. For instance, water has a specific heat of 4.186 J/g掳C, which is significantly higher than that of aluminum, which is 0.897 J/g掳C. This means that water can store more heat energy without increasing its temperature too much, compared to aluminum.

Knowing the specific heat of a material helps us calculate how much energy is needed to change its temperature by any specified amount. This calculation becomes especially important in practical applications, like designing cooling systems, where efficient energy management is crucial.
Thermal Equilibrium
Thermal equilibrium is a key concept in thermodynamics and relates to the idea that two or more objects in contact will eventually reach the same temperature. At this point, no net heat energy is exchanged between them. This balance is reached when the heat energy lost by the warmer object equals the heat energy gained by the cooler one.

In the case of the aluminum cup and water mentioned in the exercise, they are initially in thermal equilibrium at 80掳C, meaning both the cup and water have stabilized at this temperature and cease any further heat exchange.

Whenever they start being cooled uniformly, they lose energy at the same rate due to their equal temperature alignment. Calculating the rate at which energy is removed relies on the principles of thermal equilibrium, with all components contributing to the total heat loss.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In heat transfer scenarios, we often see the conversion of thermal energy (heat) into mechanical energy, electrical energy, or other forms. For instance, in the given exercise, the decrease in temperature signifies a loss of thermal energy.

The rate of this energy conversion can be expressed in watts, which is a measure of energy transfer over time, specifically joules per second. In this context, as the cup and water are cooled, the thermal energy loss is calculated and expressed as power, indicating how much energy per unit of time is being "removed" from the system.
  • The calculated loss of energy totals 5302.7 Joules per minute.
  • Since the loss must be expressed in watts, a conversion is necessary using the formula \(-88.38 \, \text{W}\).
Understanding how energy is converted and measured is essential for evaluating the efficiency of thermal processes.

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Most popular questions from this chapter

18 Overall, \(80 \%\) of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation. evaporation of sweat (2 \(430 \mathrm{~kg} / \mathrm{kg})\), evaporation from the hangs \((38 \mathrm{~kJ} / \mathrm{h})\), conduction, and convection. A person working out in a gym has a metabolic rate of \(2500 \mathrm{~kJ} / \mathrm{h} .\) His body temperature is \(37^{\circ} \mathrm{C}\), and the outside temperature \(24^{\circ} \mathrm{C}\). Assume the skin has an area of \(2.0 \mathrm{~m}^{2}\) and cmissivity of \(0.97\). (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates \(0.40 \mathrm{~kg}\) of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) AL what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

1\. A \(55-\mathrm{kg}\) woman cheats on her diet and eats a 540 Calorie (540 kcal) jelly donut for breakfast. (a) How many joules of energy are the equivalent of one jelly doughnut? (b) How many stairs must the woman climb to perform an amount of mechanical work equivalent to the food energy in one jelly doughnut? Assume the height of a single stair is \(15 \mathrm{~cm}\). (c) If the human body is only \(25 \%\) efficient in converting chemical energy to mechanical energy, how many stairs must the woman climb to work off her breakfast?

GP Into a \(0.500-\mathrm{kg}\) aluminum container at \(20.0^{\circ} \mathrm{C}\). is placed \(6.00 \mathrm{~kg}\) of ethyl alcohol at \(30.0^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~kg}\) ice at \(-10.0^{\circ} \mathrm{C}\). Assume the system is insulated from its environment. (a) Identify all five thermal energy uransfers that occur as the system goes to a final equilibrium temperature \(T\). Use the form "substance at \(\mathrm{X}^{\circ} \mathrm{C}\) to substance at \(Y^{\circ} \mathrm{C}_{.}^{\prime \prime}(\mathrm{b})\) Construct a table similar to the table in Example \(11.6 .\) (c) Sum all terms in the right-most column of the table and set the sum equal to zero. (d) Substitute information from the table into the equation found in part (c) and solve for the final equilibrium temperature, \(T\).

A class of 10 students taking an exam has a power output per student of about \(200 \mathrm{~W}\). Assume the initial temperature of the room is \(20^{\circ} \mathrm{C}\) and that its dimensions are \(6.0 \mathrm{~m}\) by \(15.0 \mathrm{~m}\) by \(3.0 \mathrm{~m}\). What is the temperature of the room at the end of \(1.0 \mathrm{~h}\) if all the energy remains in the air in the room and none is added by an outside source? The specific heat of air is \(837 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and its density is about \(1.3 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^{3}\)

A wood stove is used to heat a single room. The stove is cylindrical in shape, with a diameter of \(40.0 \mathrm{~cm}\) and a length of \(50.0 \mathrm{~cm}\), and operates at a temperature of \(400^{\circ} \mathrm{F}\). (a) If the temperature of the room is \(70.0^{\circ} \mathrm{F}\), determine the amount of radiant energy delivered to the room by the stove each second if the emissivity is \(0.920 .\) (b) If the room is a square with walls that are \(8.00 \mathrm{ft}\) high and \(25.0\) ft wide, determine the \(R\) -value needed in the walls and ceiling to maintain the inside temperature at \(70.0^{\circ} \mathrm{F}\) if the outside temperature is \(32.0^{\circ} \mathrm{F}\). Note that we are ignoring any heat conveyed by the stove via convection and any energy lost through the walls (and windows!) via convection or radiation.
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