Question

An apple with an average mass of \(0.18 \mathrm{kg}\) and average specific heat of \(3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\) is cooled from \(22^{\circ} \mathrm{C}\) to \(5^{\circ} \mathrm{C} .\) The amount of heat transferred from the apple is \((a) 0.85 \mathrm{kJ}\) (b) \(62.1 \mathrm{kJ}\) \((c) 17.7 \mathrm{kJ}\) \((d) 11.2 \mathrm{kJ}\) \((e) 7.1 \mathrm{kJ}\)

Short answer

The average mass of the apple is \(0.18 \mathrm{kg}\), and its specific heat capacity is \(3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\). Choose the closest answer among the given options. a) 7.0 kJ b) 9.0 kJ c) 10.0 kJ d) 11.2 kJ e) 12.5 kJ Answer: d) 11.2 kJ

Step-by-step solution

We have the mass of the apple (\(m = 0.18 \mathrm{kg}\)), the specific heat capacity of the apple (\(c = 3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\)), and the initial (\(T_i = 22^{\circ} \mathrm{C}\)) and final temperatures (\(T_f = 5^{\circ} \mathrm{C}\)). #Step 2: Calculate the change in temperature#

In order to calculate the amount of heat transferred, we first need to find the change in temperature, which is \(\Delta T = T_f - T_i\). Plugging in the given values, we have: \(\Delta{T} = 5^{\circ}\mathrm{C} - 22^{\circ}\mathrm{C} = -17^{\circ}\mathrm{C}\). #Step 3: Calculate the amount of heat transferred#

Using the formula for heat transfer, we have \(Q = mc\Delta T\). Plugging in the given values and the value we found for the change in temperature, we get: \(Q = (0.18 \mathrm{kg}) (3.65 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}) (-17^{\circ} \mathrm{C}) = 11.124 \mathrm{kJ}\). #Step 4: Choose the correct answer#

Comparing the result with the given options, we see that the closest answer is \(11.2\ \mathrm{kJ}\), which corresponds to option (d). Therefore, the correct answer is (d) \(11.2 \mathrm{kJ}\).

Key concepts

Specific Heat Capacity

Understanding the specific heat capacity of a substance is vital when studying energy changes within physical systems. Specific heat capacity (\(c\)) is defined as the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius (\(1^\text{o}C\) or Kelvin). This property is intrinsic to every material and helps determine how a substance will respond to heat transfer.

Take for instance our problem about the apple. Its specific heat capacity is given as \(3.65 \text{kJ/kg}^\text{o}C\) which means that every kilogram of the apple's mass requires 3.65 kJ of energy to change its temperature by one degree Celsius. This is a relatively high value when compared to metals, making the apple less susceptible to temperature change with the same amount of energy input or output.

Understanding specific heat capacity is useful not only in solving thermodynamic problems but also in real-world applications such as designing cooling systems for electronics, where materials with high specific heat capacities are ideal for absorbing excess heat.

Temperature Change

The concept of temperature change is straightforward yet crucial. It represents the difference between the initial and final temperature of a system, denoted as \( \Delta T = T_f - T_i \) where \(T_i\) is the initial temperature and \(T_f\) is the final temperature. The resulting value can be positive, indicating a gain of heat, or negative, indicating a release of heat from the system.

In our apple example, the temperature change is calculated from \(22^\text{o}C\) to \(5^\text{o}C\), resulting in a \( \Delta T\) of \( -17^\text{o}C\). A negative \( \Delta T\) confirms that the apple is indeed losing heat. This temperature change, when multiplied by the specific heat capacity and the mass of the apple, allows for the calculation of the total heat transferred. It's crucial for students to understand the significance of correctly determining \( \Delta T\) as it influences the calculation of heat transfer.

Heat Transfer Calculation

Calculating heat transfer (\(Q\)) is an integral part of thermodynamics that determines the amount of heat energy moved into or out of a system. The formula \( Q = mc\Delta T \) encapsulates the relationship between heat transfer, mass (\(m\)), specific heat capacity (\(c\)), and temperature change (\(\Delta T\)).

When an apple cools, as seen in our exercise, the negative \(\Delta T\) indicates the release of heat. By plugging in the values into the heat transfer formula, one can find out how much energy is lost during the cooling process. It is this calculation that enables the selection of the correct answer from multiple-choice questions. The correct interpretation of the calculated value of heat transferred, guided by an understanding of specific heat capacity and temperature change, leads to a successful resolution of the problem.

Through such exercises, students learn the importance of accuracy in thermodynamics and gain practical skills for real-world engineering and environmental problems. The ability to calculate heat transfer is not only a key competency in scientific disciplines but also essential for energy management and sustainability efforts.