Question

After absorbing \(1.850 \mathrm{kJ}\) of energy as heat, the temperature of a 0.500 -kg block of copper is \(37^{\circ} \mathrm{C} .\) What was its initial temperature?

Short answer

The initial temperature of the copper block was 27.38°C.

Step-by-step solution

Understanding the problem

We have a 0.500 kg block of copper absorbing 1.850 kJ of energy as heat. After absorbing this energy, the temperature of the block reaches 37°C. We need to calculate the initial temperature of the copper block.

Setting up the equation

To find the initial temperature, we use the formula \( q = mc\Delta T \), where \( q \) is the heat absorbed, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. The specific heat capacity of copper, \( c \), is 0.385 J/g°C or 385 J/kg°C.

Converting units

The heat absorbed needs to be in Joules for our equation. Convert 1.850 kJ to Joules: \( 1.850 \text{ kJ} = 1850 \text{ J} \).

Calculating \( \Delta T \)

Rearrange the formula to solve for \( \Delta T \):\[ \Delta T = \frac{q}{mc} \]Substituting the known values gives:\[ \Delta T = \frac{1850}{0.500 \times 385} \]

Performing the calculations

Calculate \( \Delta T \):\[ \Delta T = \frac{1850}{192.5} \approx 9.62 \text{ °C} \]

Finding the initial temperature

The change in temperature \( \Delta T \) is the difference between the final temperature and the initial temperature. Hence,\[ \Delta T = T_{final} - T_{initial} \]Given \( T_{final} = 37°C \), we solve for \( T_{initial} \):\[ T_{initial} = 37 - 9.62 = 27.38 \text{ °C} \]

Key concepts

Heat Transfer

Heat transfer is the process of energy moving from one object or substance to another. It typically occurs due to a temperature difference. In our exercise, the copper block absorbs energy as heat, leading to a change in its temperature.

There are three common mechanisms of heat transfer:

• Conduction: Heat transfer through direct contact, primarily occurring in solids. Copper, being a metal, efficiently conducts heat via the vibration and movement of its atoms.
• Convection: Heat transfer through the movement of fluids, which doesn't apply here.
• Radiation: Transfer of energy through electromagnetic waves, which is not significant in this context.

The absorbed heat in this scenario is accounted for by the formula \( q = mc\Delta T \). Here, \( q \) is the heat absorbed, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

By understanding this formula, we can see how heat input changes the temperature of a substance.

Temperature Change

Temperature change is a core part of the exercise and it refers to the difference in temperature a substance experiences after absorbing or losing heat. In our problem, the copper block's final temperature was provided as \( 37^{\circ} \, \mathrm{C} \).

The equation for temperature change \( \Delta T \) can be rearranged from the basic formula for heat transfer \( q = mc\Delta T \) to solve for \( \Delta T \):\[ \Delta T = \frac{q}{mc} \]Inserting the given values allows us to compute the change in temperature. Here, \( q \) was \( 1850 \, \text{J} \), \( m \) was \( 0.500 \, \text{kg} \), and \( c \), the specific heat capacity for copper, was \( 385 \, \text{J/kg°C} \).

Ultimately, the temperature change \( \Delta T \) of approximately \( 9.62 \, \text{°C} \) helped us determine that the initial temperature of the copper block was approximately \( 27.38 \, \text{°C} \) before heat was added.

Copper Properties

Copper is known for its excellent thermal and electrical conductivity, being one of the best conductors among the metals. This is due to its atomic structure, allowing free electrons to move easily and transfer energy effectively.

Here are some specific properties of copper related to the exercise:

• Low Specific Heat Capacity: At \( 385 \, \text{J/kg°C} \), it means copper doesn’t take much energy to raise its temperature compared to substances with higher heat capacities.
• High Conductivity: Its atomic structure allows heat to easily pass through, making temperature changes quick and efficient.
• Stable and Durable: Retains its properties under typical conditions and doesn’t react easily with the environment.

Because of these properties, copper is widely used in applications where efficient heat transfer is necessary, such as in cookware and heat exchangers. Understanding these characteristics helps explain why the copper block in our exercise could effectively absorb energy and change temperature.