Problem 10 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

The initial temperature was approximately \(27.38^{\circ} \text{C}.\)

Step 1: Identify the Known Values

We know the mass \( m = 0.500 \) kg, the final temperature \( T_f = 37^{\circ} \text{C} \), and the energy absorbed as heat \( Q = 1.850 \text{ kJ} = 1850 \text{ J} \). The specific heat capacity of copper is \( c = 385 \text{ J/kg}\cdot\text{°C} \).

Step 2: Use the Heat Transfer Formula

The formula for heat transfer is given by \( Q = mc\Delta T \), where \( \Delta T \) is the change in temperature. Rearrange this to \( \Delta T = \frac{Q}{mc} \).

Step 3: Calculate the Temperature Change

Plug the known values into the equation: \[\Delta T = \frac{1850}{0.500 \times 385} = \frac{1850}{192.5} \approx 9.62^{\circ} \text{C}.\]

Step 4: Find the Initial Temperature

The change in temperature \( \Delta T \) is the difference between the final and initial temperatures: \( \Delta T = T_f - T_i \). Rearrange to find the initial temperature: \( T_i = T_f - \Delta T \). Substitute the values:\[T_i = 37 - 9.62 \approx 27.38^{\circ} \text{C}.\]

Key Concepts

Heat Transfer

Heat transfer is a fascinating concept rooted in the laws of thermodynamics, describing the movement of thermal energy from a hotter object to a cooler one. When energy is added to a substance in the form of heat, it can cause the temperature of the object to rise. This process is quantified in physics by the heat transfer equation, allowing us to calculate how much energy is needed to change the temperature of a specific mass of a substance.

In our exercise, heat energy was added to a copper block, resulting in a temperature change. The heat transfer formula is important here:

• The formula is given by \( Q = mc\Delta T \), where \( Q \) is the heat energy transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
• This formula helps to link these quantities and allows us to determine the temperature change a substance undergoes when a specific amount of heat is absorbed.

Temperature Change

The concept of temperature change is integral when discussing heat transfer. It indicates how much the temperature of a substance increases or decreases when heat is added or removed.

In practical terms, the temperature change \( \Delta T \) is calculated as the difference between the final temperature after heat transfer and the initial temperature. Here’s how it was applied in the example:

• From the heat transfer equation, \( \Delta T = \frac{Q}{mc} \) was used.
• Inserting our known values, the change in temperature \( \Delta T \) was found to be approximately \( 9.62^{\circ} \text{C} \).

Understanding \( \Delta T \) is crucial because it gives insight into how a substance responds to heat energy — information vital for engineering, weather forecasting, and daily life considerations.

Copper Properties

Copper, a highly conductive metal, has interesting properties that affect how it handles heat. The specific heat capacity of a material is an indicator of how much energy is needed to raise its temperature.

For copper, the specific heat capacity is \( 385 \text{ J/kg} \cdot \text{°C} \). This comparatively low figure suggests that copper heats up and cools down quickly. Copper's properties make it an excellent choice for applications requiring efficient heat exchange, such as in heating systems or heat sinks in electronics.

• Copper's stability at high temperatures also enhances its suitability in these roles.
• Due to its efficient heat transfer characteristics, knowing copper's specific heat capacity allows for precise calculations in physics and engineering tasks.

In the exercise, knowing the specific heat capacity of copper was crucial in calculating the initial temperature from a known temperature change and absorbed heat energy.