Problem 47 - 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 237 J/kg°C.

Step 1: Understand the Energy Transfer

When the hot metal is placed on ice, it transfers heat to the ice, causing some of the ice to melt. The energy lost by the metal is equal to the energy gained by the ice.

Step 2: Calculate the Energy Required to Melt Ice

The energy required to melt a given mass of ice is calculated using the formula:\[Q_{ice} = m_{ice} \times L_f\]where:- \(Q_{ice}\) is the heat absorbed by the ice.- \(m_{ice} = 3.54\,\text{g} = 0.00354\,\text{kg}\) is the mass of ice melted.- \(L_f = 334\,\text{J/g} = 334000\,\text{J/kg}\) is the latent heat of fusion for ice.Thus, \(Q_{ice} = 0.00354\,\text{kg} \times 334000\,\text{J/kg} = 1183.16\,\text{J}\).

Step 3: Apply the Principle of Energy Conservation

According to energy conservation, the energy lost by the silver is equal to the energy gained by the ice, thus:\[Q_{silver} = Q_{ice} = 1183.16\,\text{J}\]

Step 4: Determine Specific Heat Capacity of Silver

The energy lost by the silver is calculated using the formula:\[Q_{silver} = m_{silver} \times c \times \Delta T\]where:- \(m_{silver} = 50.0\,\text{g} = 0.0500\,\text{kg}\) is the mass of the silver.- \(c\) is the specific heat capacity of silver.- \(\Delta T = 99.8\,^{\circ}\text{C} - 0.0^{\circ}\text{C} = 99.8\,^{\circ}\text{C}\) is the change in temperature.Rearranging the formula for \(c\), we have:\[c = \frac{Q_{silver}}{m_{silver} \times \Delta T} = \frac{1183.16\,\text{J}}{0.0500\,\text{kg} \times 99.8\,^{\circ}\text{C}}= \frac{1183.16}{4.99} = 237.09\,\text{J/kg}^{\circ}\text{C}\].So, the specific heat capacity of silver is approximately 237 \(\text{J/kg}^{\circ}\text{C}\).

Key Concepts

Ice Calorimeter

An ice calorimeter is a clever tool used in experiments to measure heat transfer. It involves ice, which absorbs heat when it melts. This tool is particularly useful when you want to determine how much energy a hot object, like a metal, can transfer to the ice. The main principle is simple:

• When a heated object is placed in contact with ice, it releases heat.
• The released heat causes the ice to melt.
• By measuring how much ice melts, you can figure out the amount of energy transferred.

This makes ice calorimeters a great choice in experiments to find specific heat capacities of metals.

Latent Heat of Fusion

Latent heat of fusion is a special term used in physics to describe the energy needed to change 1 kilogram of a substance from solid to liquid, without any change in temperature. For water, this value is high, at about 334,000 Joules per kilogram. This means:

• To melt 1 kg of ice at 0°C into water, 334,000 Joules of energy is required.
• This energy is involved in breaking the bonds between ice molecules.

In our exercise, the latent heat of fusion is key to calculating how much energy the melted ice absorbed from the hot silver. By using this constant, we can find out how much energy was required to melt a specific mass of ice.

Energy Conservation

The principle of energy conservation is central in physics, stating that energy cannot be created or destroyed. It can only change forms or be transferred from one object to another. In our context:

• The energy (heat) lost by the silver is exactly equal to the energy (heat) gained by the melting ice.
• This principle helps us determine the exchange of thermal energy during the experiment.

Using energy conservation, we calculated that the energy transferred from the silver heated to 99.8°C to the ice to melt 3.54 grams of it was 1183.16 Joules.

Thermal Energy Transfer

Thermal energy transfer describes how heat flows from a hotter object to a cooler one until their temperatures equalize. In the provided exercise:

• Silver at a high temperature transfers its thermal energy to the cold ice.
• This heat flow continues until the silver drops to 0°C, and the ice absorbs sufficient heat to melt.

The specific heat capacity of a substance tells us how much energy is needed to raise the temperature of 1 kilogram of the substance by 1°C. In our exercise, this concept allowed us to calculate the specific heat capacity of silver after knowing how much ice melted and the change in temperature of the metal.