Problem 4 - General Chemistry: Principles and Modern Applications

Question

A 75.0 g piece of \(\mathrm{Ag}\) metal is heated to \(80.0^{\circ} \mathrm{C}\) and dropped into \(50.0 \mathrm{g}\) of water at \(23.2^{\circ} \mathrm{C} .\) The final temperature of the \(\mathrm{Ag}-\mathrm{H}_{2} \mathrm{O}\) mixture is \(27.6^{\circ} \mathrm{C}\). What is the specific heat of silver?

Step-by-Step Solution

Answer

The specific heat of silver is \(0.235 J/g°C\).

Step 1: Calculate the Heat Gained by Water

First, let's calculate the heat gained by water using the formula \( q = mc\Delta T \). Here the mass \( m \) is \(50.0 g\), \( c \) (the specific heat capacity of water) is \( 4.18 J/g°C \), and \(\Delta T \) (the change in temperature) is \( 27.6°C - 23.2° C = 4.4° C \). Substituting these values into the equation, we find \( q = mc\Delta T = 50.0 g \times 4.18 J/g°C \times 4.4 °C = 921.92 J \). This is the heat gained by the water.

Step 2: Calculate the Heat Lost by silver

Next, understand that the heat gained by the water is equal to the heat lost by silver. Thus, the heat lost by the silver, which we'll denote by \( q_{Ag} \), is also \(921.92 J\).

Step 3: Calculate the Specific Heat of Silver

Finally, knowing the heat lost by the silver, you can calculate the specific heat of silver using the same formula \( q = mc\Delta T \). This time you know \( q_{Ag} = 921.92 J\), \( m = 75.0 g \), and \( \Delta T = 80°C - 27.6°C = 52.4°C \). Plugging these values in the equation and solving for \( c \), you get: \( c = q / m\Delta T = 921.92 J / (75.0 g \times 52.4°C ) = 0.235 J/g°C \). The specific heat of silver is therefore \( 0.235 J/g°C \).

Key Concepts

Heat Transfer

Heat transfer is the process where energy in the form of heat moves from a hotter object to a cooler one. This happens naturally and can involve various methods such as conduction, convection, and radiation. In the context of our exercise with silver and water, conduction is the method of heat transfer. When the hot silver piece is dropped into cooler water, energy transfers from the silver to the water.
This continues until both reach the same temperature, known as thermal equilibrium. During this process:

• The silver transfers heat because it initially has a higher temperature.
• Water absorbs this heat, leading to an increase in its temperature.
• The final temperature, once equilibrium is reached, is 27.6°C.

This is an ongoing exchange governed by the principle of energy flow from hot to cold until balance is achieved.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It allows us to understand energy changes within a system. The concept is applied using devices called calorimeters, which are insulated to minimize energy exchange with surroundings.
In our exercise, we don't use an actual calorimeter, but we employ calorimetry principles:

• The water and silver together can be considered a simple calorimeter system.
• The heat lost by the silver (the hotter body) is equal to the heat gained by the water (the cooler body).
• By accurately measuring temperature changes, we can calculate how much heat was transferred.

This practical application helps us determine properties like specific heat without a full calorimeter setup.

Energy Conservation

Energy conservation is a pivotal principle in physics, stating that energy cannot be created or destroyed, only transferred or converted from one form to another. In our exercise, this principle plays a crucial role.
When silver is placed in water, the total energy within the system remains constant.

• Silver loses a quantifiable amount of heat energy.
• Exactly this amount is what the water gains.
• Through calculations, you find that 921.92 Joules is the heat transferred between these masses.

Thus, the exercise exhibits energy conservation plainly through the equal energy exchange between silver and water, highlighting that even in dynamic processes, the energy of a closed system remains unchanged.