Question

How much heat is absorbed by a \(2.00 \times 10^{3}-{g}\) granite boulder \(({c}_{\text { granite }}=0.803 {J} /({g} \cdot^{\circ} {C}))\) as its temperature changes from 10.0ºC to 29.0ºC?

Short answer

The heat absorbed by the granite boulder as its temperature changes from \(10.0^{\circ}C\) to \(29.0^{\circ}C\) is \(30,514 J\).

Step-by-step solution

Identify the given values

We are given the following information: Mass of granite boulder (\(m\)): \(2.00 \times 10^{3} g\) Specific heat capacity of granite (\(c\)): \(0.803 J/g^{\circ}C\) Initial temperature (\(T_i\)): \(10.0^{\circ}C\) Final temperature (\(T_f\)): \(29.0^{\circ}C\)

Calculate the change in temperature

We need to find the change in temperature, \(\Delta T\). This can be calculated by subtracting the initial temperature from the final temperature: \(\Delta T = T_f - T_i\) Substitute the given values: \(\Delta T = 29.0^{\circ}C - 10.0^{\circ}C\) \(\Delta T = 19.0^{\circ}C\)

Calculate the heat absorbed

Now, we can calculate the heat absorbed, \(Q\), by using the formula \(Q = mc\Delta T\). Substitute the given values and the calculated change in temperature: \(Q = (2.00 \times 10^{3} g) \times (0.803 \frac{J}{g \cdot^{\circ} C}) \times (19.0^{\circ}C)\) Simplify and calculate the result: \(Q = (2.00 \times 10^{3} g) \times (0.803 J/g^{\circ}C) \times (19.0^{\circ}C)\) \(Q = (1606 J/^{\circ}C) \times (19.0^{\circ}C)\) \(Q = 30514 J\) The heat absorbed by the granite boulder as its temperature changes from \(10.0^{\circ}C\) to \(29.0^{\circ}C\) is \(30,514 J\).

Key concepts

Specific Heat Capacity

Understanding specific heat capacity is crucial when solving problems related to temperature changes in materials. It is a physical property that represents the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). Specifically, it explains why different materials heat up or cool down at different rates. For instance, the specific heat capacity of water is relatively high, which means it takes more energy to change its temperature as compared to something like granite, which has a lower specific heat capacity.

In the problem provided, we see that the specific heat capacity of granite is given as \(0.803 J/g^\text{°C}\), indicating it requires \(0.803\) joules of heat to change the temperature of one gram of granite by one degree Celsius. This value is vital for calculating the total heat absorption by the granite boulder, as we'll further explore in relation to heat transfer.

Temperature Change

When we talk about temperature change in the context of heat transfer, we refer to the difference between the final and initial temperatures, often represented as \(\text{ΔT}\). The equation for calculating temperature change is simply \(\text{ΔT} = T_f - T_i\), where \(T_f\) is the final temperature and \(T_i\) is the initial temperature. Measures of temperature change can be in degrees Celsius (°C) or Kelvin (K), but for heat calculations, both units function effectively the same because they are relative measurements.

In our exercise, the initial temperature of the granite boulder is \(10.0^\text{°C}\) and the final temperature is \(29.0^\text{°C}\). Subtracting the initial from the final gives us a temperature change of \(19.0^\text{°C}\). This information is key in calculating heat absorption, as it helps us determine how much energy has been transferred to the boulder to facilitate this change in temperature.

Heat Transfer in Thermodynamics

The concept of heat transfer in thermodynamics relates to how thermal energy is exchanged between different systems and their surroundings. There are three primary modes of heat transfer: conduction, convection, and radiation. However, in the context of our textbook example, we are dealing with the heat absorbed by a solid object, which involves conduction—the direct heat flow through a material as a result of the temperature gradient.

The equation used in the exercise, \(Q = mc\text{ΔT}\), is a foundational formula in thermodynamics for calculating the amount of heat absorbed or released during a temperature change. Here, \(Q\) denotes the heat absorbed, \(m\) is the mass of the object, \(c\) represents the specific heat capacity, and \(\text{ΔT}\) is the temperature change. By inputting the calculated temperature change of \(19.0^\text{°C}\) and the specific properties of granite, we determined the heat absorbed by the boulder is \(30,514 J\), which is a direct application of the principles of heat transfer in the field of thermodynamics.