91Ó°ÊÓ

Specific heat capacity is a crucial concept in thermodynamics. It represents the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Different substances have different specific heat capacities, which means they react to heat differently. In this problem, the specific heat is not constant, but rather varies with temperature, following the equation \(c = 0.20 + 0.14T + 0.023T^2\). This variation means that, as the temperature changes, the specific heat capacity changes too, which makes our calculation more complex. Understanding this variation is essential for accurately calculating the energy needed to change the temperature of a substance.

Energy calculation in this context typically involves determining how much energy, measured in calories or joules, is needed to change the temperature of a given mass of a substance. The formula used is derived from the principle that energy \(Q\) is equal to the mass \(m\) of the substance multiplied by the specific heat capacity \(c(T)\), integrated over the change in temperature \(T\):

• \(Q = \int_{T_1}^{T_2} mc(T) \, \mathrm{d}T\)

In our exercise, \( m = 2.0 \, \mathrm{g}\) and the temperature range is from 5°C to 15°C. By inserting the specific heat function into this integral, we are able to calculate the energy required for this temperature change. This process involves using calculus to accommodate the fact that \(c\) changes with \(T\).

Temperature change is a fundamental part of many physics and chemistry problems. Here, we're dealing with raising the temperature of a substance from 5°C to 15°C. The total temperature change \(\Delta T\) is the difference between the final and initial temperatures. In our case, \(\Delta T = 15^{\circ}\mathrm{C} - 5^{\circ}\mathrm{C} = 10^{\circ}\mathrm{C}\). Understanding this change is critical because it defines the limits of our integration for calculating energy. It's also important because the energy required is directly influenced by both the magnitude of the temperature change and how the specific heat capacity varies over that range.

Integral calculus is indispensable in physics for calculating quantities that are accumulative in nature. In this case, it allows us to add up all the tiny bits of energy needed as the temperature changes from 5°C to 15°C, where the specific heat \(c\) isn't constant. The integral

• \(\int_{5}^{15} (0.20 + 0.14T + 0.023T^2) \, \mathrm{d}T\)

is simplified by calculating each term separately:

• Constant term: \(0.20 \times \Delta T\)
• Linear term: \(0.14 \frac{T^2}{2}\)
• Quadratic term: \(0.023 \frac{T^3}{3}\)

These terms are then evaluated between the limits of 5 and 15, added together, and multiplied by the mass of the substance. This process illustrates how calculus helps handle changes across differential elements in a system.