Problem 27 - University Physics with Modern Physics

Question

An aluminum tea kettle with mass 1.50 \(\mathrm{kg}\) and containing 1.80 \(\mathrm{kg}\) 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 \(20.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C}\) ?

Step-by-Step Solution

Answer

576,874 J must be added to raise the temperature.

Step 1: Identify the Variables

First, identify the given quantities: the mass of the aluminum kettle is 1.50 kg, the mass of the water is 1.80 kg, the initial temperature is 20.0°C, and the final temperature is 85.0°C.

Step 2: Determine Specific Heat Capacities

Find the specific heat capacities: for aluminum, it is 900 J/(kg°C), and for water, it is 4186 J/(kg°C). These values are standard and widely used in heat transfer problems.

Step 3: Calculate Heat for Aluminum

Use the formula for heat, which is \( Q = mc\Delta T \). For the aluminum, it is \( Q_{\text{Al}} = 1.50 \times 900 \times (85.0 - 20.0) \). Solve this to find \( Q_{\text{Al}} = 87,750 \text{ J} \).

Step 4: Calculate Heat for Water

Similarly, calculate for water using the same formula: \( Q_{\text{Water}} = 1.80 \times 4186 \times (85.0 - 20.0) \). Solve to get \( Q_{\text{Water}} = 489,124 \text{ J} \).

Step 5: Sum the Heat Quantities

Sum the heat quantities for both aluminum and water to get the total heat required: \( Q_{\text{Total}} = Q_{\text{Al}} + Q_{\text{Water}} = 87,750 + 489,124 \). This results in \( Q_{\text{Total}} = 576,874 \text{ J} \).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property of a material that shows how much heat energy is needed to raise the temperature of one kilogram of the material by one degree Celsius. It is denoted by the letter 'c' and is measured in joules per kilogram per degree Celsius (J/kg°C). In our exercise, the specific heat capacity values are given as 900 J/(kg°C) for aluminum and 4186 J/(kg°C) for water.

These numbers mean that water needs more energy than aluminum to increase its temperature. This is because water has a higher specific heat capacity. That's why water is useful in heating systems or cooling processes. Specific heat capacity is an important factor in heat transfer calculations. Understanding this concept helps in predicting how different materials react when subjected to heat.

• Aluminum: 900 J/(kg°C)
• Water: 4186 J/(kg°C)

Heat Calculation

When it comes to calculating how much heat is needed to change the temperature of a substance, the formula we use is: \[ Q = mc\Delta T \] Here, 'Q' represents the heat energy, 'm' stands for mass, 'c' is the specific heat capacity, and \( \Delta T \) is the change in temperature.

In the context of the exercise, we have to perform this calculation twice, once for aluminum and once for water. Let's break down how this works. For aluminum, with a mass of 1.50 kg, a specific heat capacity of 900 J/(kg°C), and a temperature change from 20°C to 85°C, the calculation is:

\[ Q_{\text{Aluminum}} = 1.50 \times 900 \times (85.0 - 20.0) \] This results in 87,750 J.

• "Q" is determined by multiplying the mass, specific heat, and temperature difference.
• This calculation is similarly done for water, resulting in 489,124 J for water.
• The final step is to add up the heat for aluminum and water to find the total heat required: 576,874 J in this case.

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. When looking at our problem, it's primarily focused on the first law of thermodynamics, which concerns the conservation of energy.

In practical terms, this means that the total heat supplied to the aluminum kettle and water system is used to raise the temperature. None is assumed to be lost to the surroundings in this idealized scenario. The principle is easy enough to grasp: energy in equals energy used, assuming perfect system conditions.

• Thermodynamics helps us understand energy flow.
• It provides a framework for analyzing heating and cooling scenarios.
• Being aware of this helps solve real-world heat transfer problems effectively.

Understanding these fundamentals eases the problem-solving process in exercises like this, allowing you to predict and calculate the needed energy accurately.