/*! 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 39 A copper pot with a mass of \(0.... [FREE SOLUTION] | 91Ó°ÊÓ

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A copper pot with a mass of \(0.500 \mathrm{~kg}\) contains \(0.170 \mathrm{~kg}\) of water, and both are at \(20.0^{\circ} \mathrm{C}\). A \(0.250 \mathrm{~kg}\) block of iron at \(85.0^{\circ} \mathrm{C}\) is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Short Answer

Expert verified
The final temperature of the system, given the supplied values, is approximately \(27.9^{o}C\).

Step by step solution

01

Express heat lost

The heat lost by the iron block as it cools down is calculated using the formula \(Q = mc\Delta T\) where m is mass, c is specific heat capacity, and \(\Delta T\) is the change in temperature (initial temperature - final temperature). Given that the mass of the iron is \(0.250 \mathrm{kg}\), the initial temperature is \(85.0^{\circ} C\), finally, the specific heat capacity of iron is \(0.450 \mathrm{J/ g^{\circ} C}\). Then, convert it to \(\mathrm{J/kg^{\circ} C}\) by multiplying by 1000. The expression for the heat lost by the iron block is \(-Q_{iron} = -0.250\mathrm{kg} \cdot 450\mathrm{J/kg^{\circ} C} \cdot (85^{o}C - T_f)\)
02

Express heat gained

The heat gained by the copper pot and the water is the sum of the heats gained by each, also given by the \(Q = mc\Delta T\) where \(\Delta T\) is the final temperature - initial temperature (since each is heating up from their initial temperature to the final temperature). The mass of the water is \(0.170\mathrm{kg}\), its specific heat capacity is \(4.186 \times 10^{3} \mathrm{J/kg^{\circ} C}\) and the initial temperature is \(20.0^{o}C\). For the copper pot, the mass is \(0.500\mathrm{kg}\), the specific heat capacity is \(0.386 \times 10^{3} \mathrm{J/kg^{\circ} C}\) and the initial temperature is also \(20.0^{o}C\). The expression for the heat gained is then \(Q_{gain} = \)\[(0.170\mathrm{kg} \cdot 4.186 \times 10^{3}\mathrm{J/kg^{\circ} C} \cdot (T_f - 20^{o}C)) + (0.500\mathrm{kg} \cdot 0.386 \times 10^{3}\mathrm{J/kg^{\circ} C} \cdot (T_f - 20^{o}C))\]
03

Find the final temperature

Setting the heat lost by the iron equal to the heat gained by the pot and water, we can solve the equation for the final temperature \(T_f\). After numerous steps of algebra, this results in a quadratic equation in terms of \(T_f\). Given this equation, we can then solve for \(T_f\), giving us the final answer (using a quadratic equation solver if a calculator is available).

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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 fundamental concept in heat transfer. It describes how much energy, in the form of heat, a material needs to increase its temperature by 1°C per unit mass. Simply put, it tells us how a material reacts to adding or removing heat.
A high specific heat capacity means that a material can absorb a lot of heat without its temperature rising significantly. On the other hand, a material with a low specific heat capacity heats up quickly with less energy. This property is crucial when working out how different materials in a system will affect the final temperature during heat exchange.
In our problem, water has a specific heat capacity of 4186 J/kg°C, indicating it needs a substantial amount of energy to change temperature. Copper and iron have specific heat capacities of 386 J/kg°C and 450 J/kg°C, respectively. This means they require less heat than water to achieve the same temperature change. Understanding these differences helps us predict how a system involving water, copper, and iron will settle into a thermal equilibrium.
Thermal Equilibrium
Thermal equilibrium occurs when two or more bodies in a system reach a point where they no longer exchange heat energy. This balance is achieved when the entities reach the same temperature. In simpler terms, it's the state where everything within the system evenly distributes heat and becomes unified in terms of temperature.
In our exercise, thermal equilibrium is reached when the hot iron block and the cooler copper pot with water have transferred their heat energy back and forth until they share a common temperature. Initially, the iron block at 85°C is much hotter than the copper pot and water, which begin at 20°C. The system exchanges heat energy until the temperatures stabilize, and no further energy transfer occurs. This final temperature can be determined using the conservation of energy principles, assuming no heat is lost to the surroundings.
Conservation of Energy
The conservation of energy principle tells us energy cannot be created or destroyed, only transferred or changed from one form to another. In the context of heat transfer, this means the total amount of heat energy within a closed system remains constant, with any energy loss from one part of the system being an equal energy gain for another.
In our problem, we apply this principle to find the system's final temperature. The energy lost by the iron block ( -Q_iron), as it cools, is equal to the energy gained by the water and the copper pot (Q_gain), as they warm up. By setting these energy changes equal to each other, we can solve for the final temperature of the system. This balance ensures that all heat energy is accounted for, reflecting the universe's energy conservation rule. Understanding this principle is pivotal in solving thermal equilibrium problems, as it provides a structured approach to predicting the outcomes of energy exchanges.

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Most popular questions from this chapter

You are making pesto for your pasta and have a cylindrical measuring cup \(10.0 \mathrm{~cm}\) high made of ordinary glass \(\left[\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) that is filled with olive oil \(\left[\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right]\) to a height of \(3.00 \mathrm{~mm}\) below the top of the cup. Initially, the cup and oil are at room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\). You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{~m} \times 0.95 \mathrm{~m} \times 5.0 \mathrm{~cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8 \mathrm{~cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C},\) and the outside air temperature is \(-8.0^{\circ} \mathrm{C}\). (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window \(0.500 \mathrm{~m}\) on a side is inserted in the door? The glass is \(0.450 \mathrm{~cm}\) thick, and the glass has a thermal conductivity of \(0.80 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). The air films on the two sides of the glass have a total thermal resistance that is the same as an additional \(12.0 \mathrm{~cm}\) of glass.

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about \(1.50 \mathrm{~kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{~m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{~m} .\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

A machinist bores a hole of diameter \(1.35 \mathrm{~cm}\) in a steel plate that is at \(25.0^{\circ} \mathrm{C}\). What is the cross-sectional area of the hole (a) at \(25.0^{\circ} \mathrm{C}\) and \((\mathrm{b})\) when the temperature of the plate is increased to \(175^{\circ} \mathrm{C} ?\) Assume that the coefficient of linear expansion remains constant over this temperature range.

Animals in cold climates often depend on \(t w o\) layers of insulation: a layer of body fat (of thermal conductivity \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere \(1.5 \mathrm{~m}\) in diameter having a layer of fat \(4.0 \mathrm{~cm}\) thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at \(2.7^{\circ} \mathrm{C}\) during hibernation, while the inner surface of the fat layer is at \(31.0^{\circ} \mathrm{C}\). (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of \(50.0 \mathrm{~W} ?\) (b) How thick should the air layer (contained within the fur) be?

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