Problem 39 - Chemistry and Chemical Reactivity

Question

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 energy transferred from the metal to the ice can be determined from the amount of ice melted. Suppose you heated a 50.0 -g piece of silver to \(99.8^{\circ} \mathrm{C}\) and then dropped it onto ice. When the metal's temperature had dropped to \(0.0^{\circ} \mathrm{C},\) it is found that \(3.54 \mathrm{g}\) of ice had melted. What is the specific heat capacity of silver?

Step-by-Step Solution

Answer

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

Step 1: Understanding the Problem

We need to find the specific heat capacity of silver using the data provided from the calorimeter experiment. The amount of ice melted is related to the heat transferred from the silver to the ice, which is initially heated to a specific temperature.

Step 2: Identify Known Variables

Given: mass of silver, \( m_{Ag} = 50.0 \text{ g} \); initial temperature of silver, \( T_i = 99.8^{\circ} \text{C} \); final temperature of silver, \( T_f = 0.0^{\circ} \text{C} \); mass of ice melted, \( m_{ice} = 3.54 \text{ g} \); latent heat of fusion of ice, \( L_f = 334 \text{ J/g} \).

Step 3: Calculate Heat Absorbed by Ice

The heat absorbed by the ice can be calculated using the formula \( Q_{ice} = m_{ice} \times L_f \). Substituting the values: \( Q_{ice} = 3.54 \times 334 = 1182.36 \text{ J} \).

Step 4: Calculate Change in Temperature of Silver

The temperature change \( \Delta T \) for the silver is given by \( \Delta T = T_f - T_i = 0.0 - 99.8 = -99.8^{\circ} \text{C}\).

Step 5: Relate Heat Lost by Silver to Ice

The heat lost by the silver is equal to the heat gained by the ice: \( Q_{Ag} = Q_{ice} = 1182.36 \text{ J} \). The formula relating these is \( Q_{Ag} = m_{Ag} \times c_{Ag} \times \Delta T \), where \( c_{Ag} \) is the specific heat capacity of silver.

Step 6: Solve for Specific Heat Capacity of Silver

Rearrange the equation to solve for \( c_{Ag} \): \( c_{Ag} = \frac{Q_{Ag}}{m_{Ag} \times \Delta T} \). Substituting in the known values: \( c_{Ag} = \frac{1182.36}{50.0 \times (-99.8)} \). Calculating gives \( c_{Ag} \approx 0.237 \text{ J/g}\cdot^{\circ}\text{C} \).

Key Concepts

Ice Calorimeter

An ice calorimeter is a simple yet effective tool widely used in thermodynamic experiments to measure the specific heat capacity of materials. It operates on the principle of heat transfer, where a hot object is placed in contact with ice. As the object cools down, it transfers its heat to the ice, causing some of the ice to melt. By measuring the amount of melted ice, scientists can accurately calculate the heat exchanged during the process.
This makes ice calorimeters a practical choice in experiments due to their straightforward mechanism. It offers a precise measurement of energy transferred, which is especially useful in educational settings. The apparatus is also advantageous because it allows direct observation of the physical change (ice melting) resulting from heat transfer.

Latent Heat of Fusion

The latent heat of fusion is a crucial concept in understanding phase changes. It represents the amount of energy required to change a given mass of substance from solid to liquid at a constant temperature. For ice, this value is known to be 334 J/g. In the context of an ice calorimeter, the latent heat of fusion is used to calculate how much heat is absorbed by the melting ice.
Using the equation \( Q_{ice} = m_{ice} \times L_f \), where \( Q_{ice} \) is the heat absorbed, \( m_{ice} \) is the mass of the ice melted, and \( L_f \) is the latent heat of fusion, one can determine the total energy transferred to the ice. In experiments, knowing this value is indispensable. It allows us to accurately track the energy flow and understand how much energy is involved in melting a certain amount of ice.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics and is central in experiments involving an ice calorimeter. It refers to the process where thermal energy moves from a hotter substance to a cooler one. In the exercise, heat from the heated silver is transferred to the colder ice, thereby causing the ice to melt.
Analyzing heat transfer also helps in calculating the specific heat capacity of a material. The amount of heat loss by the silver equals the amount of heat gain by the ice, illustrating the law of conservation of energy. The relationship is expressed as \( Q_{lost} = Q_{gained} \), where \( Q_{lost} \) is the energy lost by the hot object, and \( Q_{gained} \) is the energy gained by the ice. Through this principle, students can learn how energy conservation applies in real-world scenarios.

Calorimetry Experiment

A calorimetry experiment involves measuring the heat exchanged between substances to determine properties like specific heat capacities. These experiments use devices like calorimeters to isolate the subjects from external influences, ensuring accurate measurements.
In the provided exercise, a piece of silver is heated and then placed on ice. By measuring the extent of ice melting, one can determine the energy transferred from the silver to the ice. Knowing the mass and temperature change of the silver, alongside the latent heat of fusion, enables researchers to calculate the specific heat capacity of silver. Such experiments highlight the process of energy transfer and illustrate important concepts like latent heat and conservation of energy.
Understanding these experiments helps in gaining deeper insights into thermal properties of materials, providing practical experience in handling real-world thermal dynamics.