Problem 23 - University Physics with Modern Physics

Question

An aluminum tea kettle with mass 1.10 kg and containing 1.80 kg of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from 20.0\(^\circ\)C to 85.0\(^\circ\)C?

Step-by-Step Solution

Answer

553581 J

Step 1: Understand the Specific Heat Capacity Equation

The heat required to raise the temperature of an object is given by the formula \( Q = mc\Delta T \), where \( Q \) is the heat added, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.

Step 2: Collect Known Values

The mass of the aluminum kettle \( m_1 = 1.10 \) kg and its specific heat capacity \( c_1 = 900 \) J/(kg·°C). The mass of the water \( m_2 = 1.80 \) kg and its specific heat capacity \( c_2 = 4186 \) J/(kg·°C). The initial temperature \( T_i = 20.0 \)°C, and the final temperature \( T_f = 85.0 \)°C.

Step 3: Calculate Change in Temperature

The change in temperature \( \Delta T = T_f - T_i = 85.0 \text{°C} - 20.0 \text{°C} = 65.0 \text{°C} \).

Step 4: Calculate Heat for Aluminum Kettle

For the aluminum kettle, the heat \( Q_1 = m_1c_1\Delta T = 1.10 \times 900 \times 65.0 = 64350 \text{ J} \).

Step 5: Calculate Heat for Water

For the water, the heat \( Q_2 = m_2c_2\Delta T = 1.80 \times 4186 \times 65.0 = 489231 \text{ J} \).

Step 6: Total Heat Required

The total heat required \( Q = Q_1 + Q_2 = 64350 + 489231 = 553581 \text{ J} \).

Key Concepts

Aluminum Heat Calculation

Aluminum is a commonly used metal due to its light weight and good thermal conductivity. When we want to heat aluminum, such as an aluminum tea kettle, we need to determine how much heat is needed to reach our desired temperature. This depends on its specific heat capacity, which is a measure of how much heat energy it takes to raise the temperature of a unit mass of a substance by one degree Celsius.
In our example, the specific heat capacity of aluminum is 900 J/(kg·°C). To calculate the heat required for the aluminum kettle, we use the equation:

• \( Q = mc\Delta T \)

Where:

• \( Q \) is the heat energy
• \( m \) is the mass of the aluminum (1.10 kg)
• \( c \) is the specific heat capacity (900 J/(kg·°C))
• \( \Delta T \) is the change in temperature

Inserting the values into the equation, we find:

• \( Q_1 = 1.10 \times 900 \times 65 = 64350 \) J

This calculation shows us that 64,350 Joules of energy are needed to heat the aluminum kettle from 20.0 °C to 85.0 °C.

Water Heat Calculation

Water has a higher specific heat capacity compared to many other substances, meaning it takes more energy to change its temperature. This property makes water an excellent choice for thermal applications, but it also requires more energy to heat up. The specific heat capacity of water is 4186 J/(kg·°C).
To calculate the heat required to raise the water's temperature, we use the same equation for heat energy:

• \( Q = mc\Delta T \)

For the water in our kettle, we plug in:

• \( m = 1.80 \) kg
• \( c = 4186 \) J/(kg·°C)
• \( \Delta T = 65.0 \) °C

Calculating this, we get:

• \( Q_2 = 1.80 \times 4186 \times 65 = 489231 \) J

Thus, 489,231 Joules of energy are required to raise the temperature of the water from 20.0 °C to 85.0 °C.

Temperature Change Calculation

Understanding the concept of temperature change is essential when dealing with thermal energy calculations. Temperature change is calculated as the difference between the final temperature and the initial temperature of a substance. This is represented by \( \Delta T \) in the heat capacity formula, where \( \Delta T = T_f - T_i \).
In our scenario, the temperature change for both the aluminum kettle and the water is the same:

• Initial temperature, \( T_i = 20.0 \)°C
• Final temperature, \( T_f = 85.0 \)°C

Using these values, we calculate the temperature change:

• \( \Delta T = 85.0 - 20.0 = 65.0 \)°C

By knowing \( \Delta T \), we can accurately determine the amount of heat energy required to raise the temperature for both the aluminum kettle and the water. This consistency in temperature change across different materials simplifies energy calculations because it only needs to be computed once for all components heated together.