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A 2.0 -kg metal object with a temperature of \(90^{\circ} \mathrm{C}\) is submerged in \(1.0 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C}\). The water-metal system reaches equilibrium at \(32^{\circ} \mathrm{C}\). What is the specific heat of the metal? a) \(0.840 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) b) \(0.129 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) c) \(0.512 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) b) \(0.129 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\) d) \(0.433 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\)

Short Answer

Expert verified
(Given specific heat of water is 4.186 kJ/kg. K) a) 0.055 kJ/kg K b) 0.129 kJ/kg K c) 0.2165 kJ/kg K d) 0.2585 kJ/kg K Answer: c) 0.2165 kJ/kg K

Step by step solution

01

List down the known values

We are given the following information: - Mass of metal object: \(m_\text{metal} = 2.0\) kg - Temperature of metal object: \(T_\text{metal, initial} = 90^\circ\) C - Mass of water: \(m_\text{water} = 1.0\) kg - Temperature of water: \(T_\text{water, initial} = 20^\circ\) C - Equilibrium temperature: \(T_\text{final} = 32^\circ\) C - Specific heat of water: \(c_\text{water} = 4.186 \mathrm{~ kJ/kg\cdot K}\)
02

Calculate the change in temperature for metal and water

We need to find the increase in temperature for water and the decrease in temperature for metal. For metal: $$\Delta T_\text{metal} = T_\text{final} - T_\text{metal, initial} = 32 - 90 = -58^\circ \mathrm{C}$$ For water: $$\Delta T_\text{water} = T_\text{final} - T_\text{water, initial} = 32 -20 = 12^\circ \mathrm{C}$$
03

Apply the conservation of energy equation

Using the conservation of energy equation: \(Q_{\text{metal}} = -Q_{\text{water}}\), we get, $$m_\text{metal}\times c_\text{metal}\times \Delta T_\text{metal} = -m_\text{water}\times c_\text{water}\times \Delta T_\text{water}$$
04

Solve for specific heat of metal

Now, we'll solve for \(c_\text{metal}\): $$c_\text{metal} = \frac{-m_\text{water}\times c_\text{water}\times \Delta T_\text{water}}{m_\text{metal}\times \Delta T_\text{metal}}$$ Plug in the known values: $$c_\text{metal} = \frac{-(1.0)(4.186)(12)}{(2.0)(-58)}$$ $$c_\text{metal} = 0.2165 \mathrm{~ kJ/kg\cdot K}$$ Among the given options, the closest value to our calculated specific heat is: b) \(0.129 \mathrm{~kJ} / \mathrm{kg} \mathrm{K}\)

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Key Concepts

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

Thermodynamics
Thermodynamics is a branch of physics that deals with heat, work, and temperature, and their relation to energy, radiation, and physical properties of matter. At its core, thermodynamics involves the study of energy conversions and the laws governing these transformations. One of the fundamental concepts is that energy cannot be created or destroyed; it can only be transformed from one form to another. This principle is also applied in the context of heating or cooling objects, where the energy in form of heat is either absorbed or released by the material.

When a substance like metal is heated or cooled, its internal energy changes, which can be tracked by observing changes in its temperature. The specific heat capacity of a material is a measure of the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius or Kelvin. Specific heat plays a crucial role in thermodynamics because it helps to understand how much energy is needed for a substance to achieve a certain temperature change.
Conservation of Energy
In thermodynamics, the conservation of energy is a key principle stating that the total energy of an isolated system remains constant—it is said to be conserved over time. This principle is often referred to as the first law of thermodynamics. In the context of heating or cooling, it means that the heat lost by one body must equal the heat gained by another when no other energy transfers are occurring. This rule is especially useful when analyzing closed systems, where no energy is added or removed from the environment.

In our specific heat calculation exercise, the conservation of energy tells us that the heat lost by the metal object as it cools must be equal to the heat gained by the water as it warms up. The mathematical expression of this principle lets us set up an equation that, when solved, gives us the specific heat capacity of the metal, one of the key values needed to fully characterize the thermal properties of the substance in question.
Equilibrium Temperature
Equilibrium temperature is the temperature at which two substances in thermal contact no longer transfer heat to each other. This means they have reached a state of thermal equilibrium and are at the same temperature. At this point, the net exchange of thermal energy between the substances is zero; there is no heat flow from the warmer to the cooler substance because their temperatures are the same.

In our exercise example, the metal and the water eventually reach an equilibrium temperature after the hot metal is submerged in the cooler water. Finding the equilibrium temperature is critical for solving problems involving the exchange of thermal energy, as it marks the endpoint of the energy transfer process. Understanding how to calculate equilibrium temperature is essential for a wide array of practical applications, from designing thermal systems to predicting the outcomes of heat exchange scenarios in various environments.
Heat Transfer
Heat transfer is the process by which thermal energy moves from a higher temperature object to a lower temperature object. The study of heat transfer includes three main mechanisms: conduction, convection, and radiation. In our specific heat calculation, the interaction between the metal and water involves conduction, which is the transfer of heat through a solid material or between solid materials in contact.

During the heat transfer, the metal object releases energy to the cooler water until they both achieve the same temperature. This transfer continues until equilibrium is reached. By understanding heat transfer, we can calculate the specific heat of the metal in the exercise by using the heat lost or gained and the mass and temperature change of the substances involved. When multiplied by the change in temperature, the specific heat capacity allows us to quantify the amount of heat transferred, thus linking the concept of heat transfer with the actual change in thermal energy of the substances.

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Most popular questions from this chapter

Assuming the severity of a burn increases as the amount of energy put into the skin increases, which of the following would cause the most severe burn (assume equal masses)? a) water at \(90^{\circ} \mathrm{C}\) b) copper at \(110^{\circ} \mathrm{C}\) c) steam at \(180^{\circ} \mathrm{C}\) d) aluminum at \(100^{\circ} \mathrm{C}\) e) lead at \(100^{\circ} \mathrm{C}\)

A cryogenic storage container holds liquid helium, which boils at \(4.2 \mathrm{~K}\). Suppose a student painted the outer shell of the container black, turning it into a pseudoblackbody, and that the shell has an effective area of \(0.50 \mathrm{~m}^{2}\) and is at \(3.0 \cdot 10^{2} \mathrm{~K}\). a) Determine the rate of heat loss due to radiation. b) What is the rate at which the volume of the liquid helium in the container decreases as a result of boiling off? The latent heat of vaporization of liquid helium is \(20.9 \mathrm{~kJ} / \mathrm{kg} .\) The density of liquid helium is \(0.125 \mathrm{~kg} / \mathrm{L}\).

The \(R\) factor for housing insulation gives the thermal resistance in units of \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\). A good wall for harsh climates, corresponding to about 10.0 in of fiberglass, has \(R=40.0 \mathrm{ft}^{2}{ }^{\circ} \mathrm{F} \mathrm{h} / \mathrm{BTU}\) a) Determine the thermal resistance in SI units. b) Find the heat flow per square meter through a wall that has insulation with an \(R\) factor of 40.0 , with an outside temperature of \(-22.0^{\circ} \mathrm{C}\) and an inside temperature of \(23.0^{\circ} \mathrm{C}\)

In one of your rigorous workout sessions, you lost \(150 \mathrm{~g}\) of water through evaporation. Assume that the amount of work done by your body was \(1.80 \cdot 10^{5} \mathrm{~J}\) and that the heat required to evaporate the water came from your body. a) Find the loss in internal energy of your body, assuming the latent heat of vaporization is \(2.42 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}\). b) Determine the minimum number of food calories that must be consumed to replace the internal energy lost (1 food calorie \(=4186\) J).

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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