When \(25.0 \mathrm{~g}\) of a metal at a temperature of \(90.0^{\circ} \mathrm{C}\) is added to \(50.0 \mathrm{~g}\) of water at \(25.0^{\circ} \mathrm{C}\), the water temperature rises to \(29.8^{\circ} \mathrm{C}\). The specific heat capacity of water is \(4.184 \mathrm{~J}-\left({ }^{\circ} \mathrm{C}\right)^{-1} \mathrm{~g}^{-1}\). What is the specific heat capacity of the metal?

Short Answer

Expert verified
The specific heat capacity of the metal is approximately \(0.387 \mathrm{~J/g^\circ C}\).

Step by step solution

01

Understand the Concept of Heat Exchange

The heat lost by the metal will be equal to the heat gained by the water. This is based on the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred.
02

Calculate the Heat Gained by the Water

Use the formula for heat transfer, which is \( Q = mc\Delta T \), where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Here, \( Q_{water} = m_{water} \times c_{water} \times \Delta T_{water} \). Plug in the values for water to find \( Q_{water} \).
03

Calculate the Change in Temperature of the Metal

The change in temperature of the metal, \( \Delta T_{metal} \), can be found by subtracting the final temperature of the system from the initial temperature of the metal: \( \Delta T_{metal} = 90.0^\circ\mathrm{C} - 29.8^\circ\mathrm{C} \).
04

Set Up the Equation for Heat Lost by the Metal

Apply the heat transfer formula to the metal: \( Q_{metal} = m_{metal} \times c_{metal} \times \Delta T_{metal} \). Since the heat lost by the metal is equal to the heat gained by the water, set \( Q_{metal} \) equal to \( Q_{water} \).
05

Solve for the Specific Heat Capacity of the Metal

Rearrange the equation \( Q_{metal} = Q_{water} \) to solve for \( c_{metal} \). The formula becomes \( c_{metal} = \frac{Q_{water}}{m_{metal} \times \Delta T_{metal}} \). Use the previously calculated values for \( Q_{water} \) and \( \Delta T_{metal} \) to find \( c_{metal} \).

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics and plays a crucial role in understanding how energy moves from one object to another. Simply put, it's the process that occurs when a warmer object comes into contact with a cooler one, causing energy to flow from the former to the latter until both reach the same temperature.

Let's break it down using an everyday example: When you pour hot coffee into a cold mug, heat transfer occurs. The heat from the coffee flows to the mug until both have reached a balanced temperature. In the context of our exercise, the hot metal transfers heat to the cooler water until they both reach thermal equilibrium.

To calculate heat transfer, we use the formula:
\( Q = mc\Delta T \),
where \( Q \) is the heat transferred, \( m \) is the mass of the substance, \( c \) is its specific heat capacity, and \( \Delta T \) is the change in temperature. Understanding this formula is key to solving problems involving heat transfer, such as finding the specific heat capacity of an unknown metal when placed in water.
Thermal Equilibrium
Thermal equilibrium is a state where two objects that are in contact with each other reach the same temperature and no further heat transfer occurs between them. This is because the rate of heat transfer from the hotter to the cooler object slows down as the temperature difference between them decreases, and it eventually stops when both temperatures are equal. It's like when you stop feeling the warmth from a heating pad once your body adjusts to its temperature.

In our exercise, the metal and the water eventually reach the same temperature, which indicates they have achieved thermal equilibrium. The water's temperature stops rising because it has balanced out the temperature with the metal. The concept of thermal equilibrium plays an essential role in determining the final temperature in heat transfer exercises. It allows us to set the heat gained by the water equal to the heat lost by the metal, thereby enabling us to solve for unknown quantities like the specific heat capacity of the metal.
Conservation of Energy
The law of conservation of energy states that energy cannot be created or destroyed; it can only change forms or be transferred from one object to another. In the realm of our textbook exercise, this principle assures us that the heat lost by the metal is the same amount of heat gained by the water, as energy is simply moving from the metal to the water.

When we apply this concept to solve for the specific heat capacity of the metal, we set up an equation based on the idea that the total energy before and after the heat transfer remains constant. By knowing this, we can make the logical assumption that:
\( Q_{metal} = Q_{water} \).
This relationship allows us to find unknown variables, such as the specific heat capacity of the metal, by manipulating the heat transfer formula. This demonstrates the power of the conservation of energy principle in practical applications like determining physical properties from thermal interactions.

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Most popular questions from this chapter

Use the enthalpies of formation in Appendix \(2 \mathrm{~A}\) to calculate the standard enthalpy of the following reactions: (a) the replacement of deuterium by ordinary hydrogen in heavy water: \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})+\mathrm{D}_{2}(\mathrm{~g})\) (b) the removal of sulfur from the hydrogen sulfide and sulfur dioxide in natural gas: \(2 \mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})+\mathrm{SO}_{2}(\mathrm{~g}) \rightarrow 3 \mathrm{~S}(\mathrm{~s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) the oxidation of ammonia: \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\)

A gas sample is heated in a cylinder, using \(375 \mathrm{~kJ}\) of heat. At the same time a piston compresses the gas, using \(645 \mathrm{~kJ}\) of work. What is the change in internal energy of the gas during this process?

Calculate the reaction enthalpy for the formation of anhydrous aluminum chloride, \(2 \mathrm{Al}(\mathrm{s})+3 \mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{AlCl}_{3}(\mathrm{~s})\), from the following data: $$ \begin{array}{ll} 2 \mathrm{Al}(\mathrm{s})+6 \mathrm{HCl}(\mathrm{aq}) \longrightarrow 2 \mathrm{AlCl}_{3}(\mathrm{aq})+3 \mathrm{H}_{2}(\mathrm{~g}) \\ & \Delta H^{\circ}=-1049 \mathrm{~kJ} \\ \mathrm{HCl}(\mathrm{g}) \longrightarrow \mathrm{HCl}(\mathrm{aq}) & \Delta H^{\circ}=-74.8 \mathrm{~kJ} \\ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{HCl}(\mathrm{g}) & \Delta H^{\circ}=-185 \mathrm{~kJ} \\ \mathrm{AlCl}_{3}(\mathrm{~s}) \longrightarrow \mathrm{AlCl}_{3}(\mathrm{aq}) & \Delta H^{\circ}=-323 \mathrm{~kJ} \end{array} $$

For a certain reaction at constant pressure, \(\Delta H=-15 \mathrm{~kJ}\) and \(22 \mathrm{~kJ}\) of expansion work is done on the system. What is \(\Delta U\) for this process?

In a combustion cylinder, the total internal energy change produced from the burning of a fuel is \(-2573 \mathrm{~kJ}\). The cooling system that surrounds the cylinder absorbs \(947 \mathrm{~kJ}\) as heat. How much work can be done by the fuel in the cylinder during this process?

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