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An "ice calorimeter" can be used to determine the specific heat capacity of a metal. A piece of hot metal is dropped onto a weighed quantity of ice. The quantity of heat transferred from the metal to the ice can be determined from the amount of ice melted. Suppose you heat a 50.0 -g piece of silver to \(99.8^{\circ} \mathrm{C}\) and then drop it onto ice. When the metal's temperature has dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(3.54 \mathrm{g}\) of ice has melted. What is the specific heat capacity of silver?

Short Answer

Expert verified
The specific heat capacity of silver is approximately 0.237 J/g°C.

Step by step solution

01

Understand the Problem

We need to find the specific heat capacity of silver. We start by knowing that the heat lost by silver equals the heat gained by the ice, which causes the ice to melt. The known data: 50.0 g of silver was originally at 99.8°C, and 3.54 g of ice melted.
02

Calculate Heat Absorbed by Ice

The heat absorbed by ice can be calculated using the latent heat of fusion for ice, which is about 334 J/g. The amount of heat, \( q_{\text{ice}} \), absorbed by the ice is given by \( q_{\text{ice}} = m_{\text{ice}} \times L_f \). Thus, \( q_{\text{ice}} = 3.54 \, \text{g} \times 334 \, \text{J/g} = 1183.56 \, \text{J} \).
03

Relate Heat Lost by Silver to Heat Gained by Ice

By conservation of energy, the heat lost by silver, \( q_{\text{silver}} \), is equal to the heat gained by the ice: \( q_{\text{silver}} = q_{\text{ice}} = 1183.56 \, \text{J} \).
04

Calculate the Specific Heat Capacity of Silver

Use the formula to calculate heat, \( q = m c \Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature. For silver, \( q_{\text{silver}} = m_{\text{silver}} \times c_{\text{silver}} \times \Delta T \). We know \( q_{\text{silver}} = 1183.56 \, \text{J} \), \( m_{\text{silver}} = 50.0 \, \text{g} \), and \( \Delta T = 99.8^{\circ} \text{C} - 0^{\circ} \text{C} = 99.8 \, \text{C} \). Thus, \( c_{\text{silver}} = \frac{q_{\text{silver}}}{m_{\text{silver}} \times \Delta T} = \frac{1183.56}{50.0 \times 99.8} \approx 0.237 \, \text{J/g}^\circ \text{C} \).

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Key Concepts

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

Ice Calorimeter
An ice calorimeter is a handy tool in experiments where we aim to measure the amount of heat transferred to or from substances using the melting of ice as our measure. This calorimeter makes use of the principle that when heat is absorbed by ice, it changes state from solid to liquid, a process known as melting. This melting process is harnessed to measure the heat that has been transferred, giving us invaluable data without requiring complex equipment.

When we drop a heated metal into the ice, the heat from the metal is absorbed by the ice, specifically in melting it. This change in state from ice to water requires a precise amount of energy, thus, by measuring the mass of the ice that melts, we can calculate the heat transferred. Remember, the amount of ice melted directly relates to the energy transfer, making this device perfect for calorimetric measurements like the one described in the given problem.

The capability of the ice calorimeter to determine specific heat capacities of various substances, such as metals, demonstrates its practical efficiency in a laboratory setting.
Latent Heat of Fusion
Latent heat of fusion is the amount of energy needed to change a substance from solid to liquid at its melting point without changing its temperature. This is crucial for working with an ice calorimeter, as the latent heat of fusion tells us exactly how much energy is required to melt a certain amount of ice.

For ice, the latent heat of fusion is about 334 Joules per gram. This means that to melt just one gram of ice, 334 Joules of energy must be absorbed. In our problem, 3.54 grams of ice melted, which means the ice absorbed a significant amount of energy — specifically, 1183.56 Joules.

By knowing the latent heat of fusion, you can directly relate the mass of melted ice to the amount of heat transferred, which is why it is so vital in calorimetry experiments. In our problem, the energy absorbed by the ice in melting was equal to the energy lost by the silver, allowing for a precise calculation of silver's heat transfer and its specific heat capacity.
Conservation of Energy
The principle of conservation of energy is fundamental to understanding calorimetry experiments. It states that energy cannot be created or destroyed in an isolated system. This principle is crucial when determining heat transfer between substances, like in the example of a heated metal dropped onto ice.

In our exercise, the conservation of energy allows us to equate the heat lost by the cooling silver to the heat gained by the ice. So, the heat transferred to melt the ice must equal the heat given off by the silver as it cools down. By maintaining this balance, we can calculate unknowns such as the specific heat capacity of silver.

This ensures that all energy transfers are accounted for, highlighting the intrinsic balance in calorimetry readings and how they reflect true physical interactions. Thus, conservation of energy acts as a pivotal guide to frame and solve thermodynamic problems such as this one.
Heat Transfer
Heat transfer involves the movement of thermal energy from a hotter object to a cooler one, aiming to reach thermal equilibrium. Understanding heat transfer is essential when measuring the specific heat capacity of substances, as demonstrated in the provided problem.

When the hot silver is placed on the ice, heat from the silver is transferred to the ice, causing it to melt. The process can be quantified using the formula for heat transfer, \( q = mc\Delta T \), where \( q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

This formula allowed us to calculate how much heat was transferred from the silver to the ice, providing enough information to compute the specific heat capacity of silver. By placing these principles into practice, one can decipher the intricate details of thermal interactions and properties of materials in various environments.

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Most popular questions from this chapter

The following terms are used extensively in thermodynamics. Define each and give an example. (a) exothermic and endothermic (b) system and surroundings (c) specific heat capacity (d) state function (e) standard state (f) enthalpy change, \(\Delta H\) (g) standard enthalpy of formation

A 0.692 -g sample of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6},\) is burned in a constant volume calorimeter. The temperature rises from \(21.70^{\circ} \mathrm{C}\) to \(25.22^{\circ} \mathrm{C} .\) The calorimeter contains \(575 \mathrm{g}\) of water and the bomb has a heat capacity of \(650 \mathrm{J} / \mathrm{K}\). What quantity of heat is evolved per mole of glucose?

After absorbing \(1.850 \mathrm{kJ}\) of heat, the temperature of a \(0.500-\mathrm{kg}\) block of copper is \(37^{\circ} \mathrm{C} .\) What was its initial temperature?

Calcium carbide, \(\mathrm{CaC}_{2}\), is manufactured by the reaction of CaO with carbon at a high temperature. (Calcium carbide is then used to make acetylene.) $$\mathrm{CaO}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \longrightarrow \mathrm{CaC}_{2}(\mathrm{s})+\mathrm{CO}(\mathrm{g})$$ $$\Delta H_{\mathrm{rxn}}^{\circ}=+464.8 \mathrm{kJ}$$ Is this reaction endothermic or exothermic? If \(10.0 \mathrm{g}\) of CaO is allowed to react with an excess of carbon, what quantity of heat is absorbed or evolved by the reaction?

The standard enthalpy of formation of solid barium oxide, \(\mathrm{BaO},\) is \(-553.5 \mathrm{kJ} / \mathrm{mol},\) and the enthalpy of formation of barium peroxide, \(\mathrm{BaO}_{2},\) is \(-634.3 \mathrm{kJ} / \mathrm{mol}\). (a) Calculate the standard enthalpy change for the following reaction. Is the reaction exothermic or endothermic? $$\mathrm{BaO}_{2}(\mathrm{s}) \longrightarrow \mathrm{BaO}(\mathrm{s})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ (b) Draw an energy level diagram that shows the relationship between the enthalpy change of the decomposition of \(\mathrm{BaO}_{2}\) to \(\mathrm{BaO}\) and \(\mathrm{O}_{2}\) and the enthalpies of formation of \(\mathrm{BaO}(\mathrm{s})\) and \(\mathrm{BaO}_{2}(\mathrm{s})\).

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