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Chapter 18: Problem 7

An aluminum block of mass \(M_{\mathrm{Al}}=2.0 \mathrm{~kg}\) and specific heat \(C_{\mathrm{Al}}=910 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) is at an initial temperature of \(1000{ }^{\circ} \mathrm{C}\) and is dropped into a bucket of water. The water has mass \(M_{\mathrm{H}_{2} \mathrm{O}}=12 \mathrm{~kg}\) and specific heat \(C_{\mathrm{H}_{2} \mathrm{O}}=4190 \mathrm{~J} /(\mathrm{kg} \mathrm{K})\) and is at room temperature \(\left(25^{\circ} \mathrm{C}\right) .\) What is the approximate final temperature of the system when it reaches thermal equilibrium? (Neglect heat loss out of the system.) a) \(50^{\circ} \mathrm{C}\) b) \(60^{\circ} \mathrm{C}\) c) \(70^{\circ} \mathrm{C}\) d) \(80^{\circ} \mathrm{C}\)

Short Answer

Expert verified
a) 50 掳C

Step by step solution

01

Calculate the energy lost by the aluminum block

To calculate the energy lost by the aluminum block, we can use the formula: \(q_{Al} = M_{Al} \times C_{Al} \times (T_{initial} - T_{final})\) where \(T_{initial}\) is the initial temperature of the aluminum block (1000 掳C), and \(T_{final}\) is the final temperature we want to determine. We will rewrite the formula above, leaving \(T_{final}\) as the subject: \(T_{final} = T_{initial} - \frac{q_{Al}}{M_{Al} \times C_{Al}}\)
02

Calculate the energy gained by the water

To calculate the energy gained by the water, we can use the formula: \(q_{H_2O} = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\) where \(T_{initial}\) is the initial temperature of the water (25 掳C), and \(T_{final}\) is the final temperature we want to determine. We will rewrite the formula above, leaving \(q_{H_2O}\) as the subject: \(q_{H_2O} = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\)
03

Apply conservation of energy

As stated before, the energy lost by the aluminum block must be equal to the energy gained by the water. Therefore, we can equate the two formulas from Step 1 and Step 2: \(M_{Al} \times C_{Al} \times (T_{initial} - T_{final}) = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial})\) Now we can plug in the given values for mass and specific heat and solve for \(T_{final}\): \(2.0\ Kg \times 910\ J/(KgK) \times (1000 - T_{final}) = 12.0\ Kg \times 4190\ J/(KgK) \times (T_{final} - 25)\)
04

Solve for the final temperature

To solve for \(T_{final}\), first expand the equation and simplify: \(1820(1000 - T_{final}) = 50280(T_{final} - 25)\) \(1820000 - 1820T_{final} = 50280T_{final} - 1257000\) Now combine like terms and solve for \(T_{final}\): \(1820T_{final} + 50280T_{final} = 1820000 - 1257000\) \(52100T_{final} = 563000\) \(T_{final} = 563000 \div 52100\) \(T_{final} \approx 10.8\) Since the initial temperature of the water was 25 掳C, the final temperature is about 25 + 10.8 = 35.8 掳C, which is closest to the given option a) \(50^{\circ} \mathrm{C}.\) Thus, the approximate final temperature of the system when it reaches thermal equilibrium is 50 掳C.

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Key Concepts

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

Specific Heat
The specific heat of a substance is an inherent property that indicates the amount of heat required to change the temperature of a unit mass of the substance by one degree Celsius. It tells us how much energy a material can store. For example, aluminum and water have different specific heat values. Aluminum has a specific heat of 910 J/(kg K), meaning it requires 910 joules to raise the temperature of one kilogram of aluminum by one degree Celsius. In contrast, water has a much higher specific heat of 4190 J/(kg K), which means it can store more heat energy than aluminum. This property significantly affects how these materials behave when they exchange heat. Higher specific heat means that the substance can absorb more heat without a large change in temperature. This is why, in this exercise, the energy transfer will not cause a massive temperature increase in water compared to aluminum's temperature decrease.
Energy Conservation
Energy conservation in physics refers to the principle that energy in a closed system remains constant, regardless of the processes happening within the system. This principle is the backbone of solving problems like the one in the exercise. Here, the energy that the aluminum block loses must equal the energy that the water gains, assuming no heat loss to the surroundings. In mathematical terms, this is expressed as:
  • The heat lost by aluminum = The heat gained by water
This is written as:\[ M_{Al} \times C_{Al} \times (T_{initial, Al} - T_{final}) = M_{H_2O} \times C_{H_2O} \times (T_{final} - T_{initial, H_2O}) \]This equation allows us to solve for the unknown final temperature, making it a powerful tool for analyzing the thermal interactions between substances.
Heat Transfer
Heat transfer is the process through which thermal energy moves from one object or material to another. In our example, the mechanism of heat transfer is conduction, as thermal energy is exchanged directly between the aluminum block and the water. When the system is not experiencing any external heat loss, as in this exercise, the heat from the aluminum (which is initially at a higher temperature) will flow into the water (initially at a lower temperature) until thermal equilibrium is reached, meaning both substances arrive at a common temperature. Several factors influence the rate and amount of heat transfer:
  • The temperature difference between the materials: Greater differences increase the heat transfer rate.
  • The surface area in contact: Larger contact area allows more heat to pass through.
  • The specific heat of the materials: As discussed, higher specific heat means a material will absorb more energy without a large temperature change.
Understanding heat transfer is crucial to predicting how different materials interact thermally when brought into contact.

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Most popular questions from this chapter

You are going to lift an elephant (mass \(\left.=5.0 \cdot 10^{3} \mathrm{~kg}\right)\) over your head \((2.0 \mathrm{~m}\) vertical displacement). a) Calculate the work required to do this. You will lift the elephant slowly (no tossing of the elephant allowed!). If you want, you can use a pulley system. (As you saw in Chapter \(5,\) this does not change the energy required to lift the elephant, but it definitely reduces the force required to do so.) b) How many doughnuts ( 250 food calories each) must you metabolize to supply the energy for this feat?

Why might a hiker prefer a plastic bottle to an old-fashioned aluminum canteen for carrying his drinking water?

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Arthur Clarke wrote an interesting short story called "A Slight Case of Sunstroke." Disgruntled football fans came to the stadium one day equipped with mirrors and were ready to barbecue the referee if he favored one team over the other. Imagine the referee to be a cylinder filled with water of mass \(60.0 \mathrm{~kg}\) at \(35.0^{\circ} \mathrm{C}\). Also imagine that this cylinder absorbs all the light reflected on it from 50,000 mirrors. If the heat capacity of water is \(4.20 \cdot 10^{3} \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right),\) how long will it take to raise the temperature of the water to \(100 .{ }^{\circ} \mathrm{C}\) ? Assume that the Sun gives out \(1.00 \cdot 10^{3} \mathrm{~W} / \mathrm{m}^{2},\) the dimensions of each mirror are \(25.0 \mathrm{~cm}\) by \(25.0 \mathrm{~cm},\) and the mirrors are held at an angle of \(45.0^{\circ}\)

A thermal window consists of two panes of glass separated by an air gap. Each pane of glass is \(3.00 \mathrm{~mm}\) thick, and the air gap is \(1.00 \mathrm{~cm}\) thick. Window glass has a thermal conductivity of \(1.00 \mathrm{~W} /(\mathrm{m} \mathrm{K})\), and air has a thermal conductivity of \(0.0260 \mathrm{~W} /(\mathrm{m} \mathrm{K})\). Suppose a thermal window separates a room at temperature \(20.00{ }^{\circ} \mathrm{C}\) from the outside at \(0.00^{\circ} \mathrm{C}\). a) What is the temperature at each of the four air-glass interfaces? b) At what rate is heat lost from the room, per square meter of window? c) Suppose the window had no air gap but consisted of a single layer of glass \(6.00 \mathrm{~mm}\) thick. What would the rate of heat loss per square meter be then, under the same temperature conditions? d) Heat conduction through the thermal window could be reduced essentially to zero by evacuating the space between the glass panes. Why is this not done?
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