Problem 99 - General Chemistry

Question

A piece of lead of mass \(121.6 \mathrm{~g}\) was heated by an electrical coil. From the resistance of the coil, the current, and the time the current flowed, it was calculated that \(235 \mathrm{~J}\) of heat was added to the lead. The temperature of the lead rose from \(20.4^{\circ} \mathrm{C}\) to \(35.5^{\circ} \mathrm{C}\). What is the specific heat of the lead?

Step-by-Step Solution

Answer

The specific heat of the lead is approximately 0.128 J/g°C.

Step 1: Understand the problem

We need to find the specific heat capacity of lead given the mass, amount of heat added, and change in temperature.

Step 2: Set up the formula

The formula to calculate specific heat capacity is given by: \[ q = mc\Delta T \] where \( q \) is the heat added (in Joules), \( m \) is the mass (in grams), \( c \) is the specific heat capacity (in \( \text{J/g}^{\circ} \text{C} \)), and \( \Delta T \) is the change in temperature (in degrees Celsius).

Step 3: Substitute the known values into the formula

From the problem, we know: \( q = 235 \text{ J} \), \( m = 121.6 \text{ g} \), and \( \Delta T = 35.5^{\circ} \text{C} - 20.4^{\circ} \text{C} = 15.1^{\circ} \text{C} \). Substitute these values into the formula: \[ 235 = 121.6 \times c \times 15.1 \]

Step 4: Solve for specific heat capacity

We solve the equation \( 235 = 121.6 \times c \times 15.1 \) to find \( c \). Rearrange the equation to isolate \( c \): \[ c = \frac{235}{121.6 \times 15.1} \] Calculate the value of \( c \).

Step 5: Calculate the result

Perform the division: \[ c = \frac{235}{1835.36} \approx 0.128 \text{ J/g}^{\circ} \text{C} \].

Key Concepts

Thermal Energy

Thermal energy is a type of energy that is associated with the temperature of an object. It is the total kinetic energy of the particles in a substance. When a substance is heated, its particles move faster, leading to an increase in thermal energy.
Understanding thermal energy is crucial as it is the basis for heat calculations and temperature changes. In our exercise, the lead piece absorbed thermal energy from an electrical coil, causing its temperature to rise. This process is common in many real-world situations, where heat is transferred to substances to change their energy state or enhance certain properties.
In practical terms, thermal energy can be added or removed to induce temperature changes, which are often calculated using specific formulas. The amount of thermal energy transferred is measured in Joules (J), which is a standard unit in physics.
Whenever you encounter a problem involving thermal energy, think about how the energy affects the object’s state and how you can measure or calculate this change.

Temperature Change

Temperature change is a measure of how much the temperature of a substance increases or decreases. In our problem, the temperature of the lead piece rose from 20.4°C to 35.5°C, making a temperature change of 15.1°C.
When calculating temperature change, use the formula:

• \( \Delta T = T_{final} - T_{initial} \)

Where \( T_{final} \) is the final temperature, and \( T_{initial} \) is the starting temperature.
This concept is key when determining the effects of heat addition. The change in temperature directly relates to how much thermal energy is absorbed.
A larger temperature increase usually means more energy absorption, while a smaller change may indicate less energy was introduced. Assessing temperature change helps us determine the effectiveness of heating or cooling methods in various applications.

Heat Calculation

Calculating heat involves determining how much energy is needed to change a substance's temperature. The formula used is:

• \( q = mc\Delta T \)

Here, \( q \) is the heat added in Joules, \( m \) is the object’s mass, \( c \) is the substance's specific heat capacity, and \( \Delta T \) is the temperature change.
In our example, we wanted to find the specific heat capacity of lead. We already knew the heat added, the mass of the lead, and the change in temperature, making \( c \) the unknown factor.
By rearranging the formula to solve for \( c \):

• \( c = \frac{q}{m \times \Delta T} \)

We derived the specific heat by plugging in the known values, allowing us to understand how lead responds to added heat.
Performing such heat calculations is vital in fields ranging from materials science to everyday cooking, ensuring predictable temperature control and energy use.