Problem 29 - College Physics

Question

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, which is then transferred to the liquid for \(120 \mathrm{~s}\) at a constant rate of \(65.0 \mathrm{~W}\). The mass of the liquid is \(0.780 \mathrm{~kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\). (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or its surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

Step-by-Step Solution

Answer

(a) The specific heat is approximately 2610.77 J/(kg·°C). (b) It's an underestimate due to unaccounted heat loss.

Step 1: Calculate the total energy supplied

The electrical power supplied is given by \( P = 65.0 \, \text{W} \). The power is supplied for a time \( t = 120 \, \text{s} \). Therefore, the total energy \( Q \) transferred to the liquid is calculated as follows:\[ Q = P \times t = 65.0 \, \text{W} \times 120 \, \text{s} = 7800 \, \text{J} \]

Step 2: Calculate the temperature change of the liquid

The initial and final temperatures of the liquid are \( T_i = 18.55^{\circ} \text{C} \) and \( T_f = 22.54^{\circ} \text{C} \) respectively. Therefore, the change in temperature \( \Delta T \) is:\[ \Delta T = T_f - T_i = 22.54 - 18.55 = 3.99^{\circ} \text{C} \]

Step 3: Calculate the specific heat capacity

The specific heat capacity \( c \) can be calculated using the formula:\[ Q = m \cdot c \cdot \Delta T \]Where:- \( m = 0.780 \, \text{kg} \) (mass of the liquid)- \( \Delta T = 3.99^{\circ} \text{C} \) (temperature change)Rearranging for \( c \):\[ c = \frac{Q}{m \cdot \Delta T} = \frac{7800 \, \text{J}}{0.780 \, \text{kg} \times 3.99^{\circ} \text{C}} \approx 2610.77 \, \text{J/(kg} \cdot \text{°C)} \]

Step 4: Analyze possible heat losses

If heat transfer from the liquid to the container or its surroundings is considered, the calculated specific heat would be an underestimate. This is because some of the energy would dissipate into the surroundings instead of transferring into the liquid, leading to a higher energy requirement to achieve the measured temperature change.

Key Concepts

Heat Transfer

Heat transfer is a fundamental concept in physics and refers to the movement of thermal energy from one object or medium to another. In the experiment described, heat transfer occurs when the energy generated by the electrical resistor is transferred to the liquid. This transfer of energy causes the liquid's temperature to rise.

It is crucial to understand that heat transfer can happen via three main mechanisms:

• Conduction: Direct transfer of heat through a material.
• Convection: Transfer of heat through a fluid, including gases and liquids, by the movement of the fluid itself.
• Radiation: Transfer of heat through electromagnetic waves.

In our case, it's a form of conduction and possibly convection as the heat generated in the resistor spreads through the liquid. The efficiency of heat transfer directly influences the effectiveness of energy use in various applications, including cooking, heating systems, and industrial processes.

Energy Conversion

Energy conversion is the process of transforming energy from one form to another. In the context of this experiment, electrical energy from the resistor is converted into thermal energy or heat. This is a common phenomenon where energy in the form of electricity is used to produce heat, as seen in devices like electric stoves and water heaters.

This conversion is pivotal when calculating the energy supplied to a system. The power of the resistor, given in watts (a unit of power equivalent to joules per second), helps calculate the total energy converted to heat over a specific duration. It is important to note that during energy conversion, some energy might be lost to the surroundings, becoming important in determining the experiment's efficiency. Special attention must be paid to ensure that all energy conversions are accounted for, as any unacknowledged energy loss can lead to errors in calculations, affecting factors such as the specific heat capacities of substances involved.

Temperature Change

Temperature change is a measure of how much the thermal energy of a substance has increased or decreased. In the described experiment, the identification of temperature change allows us to compute the specific heat capacity of the unknown liquid.

The change in temperature (\( \Delta T \)) is calculated simply by subtracting the initial temperature from the final temperature. This simple but crucial calculation is essential as it helps us understand how the liquid absorbs heat. If the temperature of the liquid increases significantly with a small amount of heat, it indicates a low specific heat capacity. Conversely, a large amount of energy causing only a small change in temperature signifies a high specific heat capacity.Temperature change can be affected by surrounding conditions, emphasizing the need for careful control of heat loss during experiments. Variations can lead to inaccurate specific heat capacity readings in practical applications.

Experimental Physics

Experimental physics involves the practical side of analyzing and understanding the laws of physics through experimentation and observation. The scenario in this task represents a classic type of laboratory experiment aimed at measuring specific heat capacity.

In experimental physics, controlling variables is essential. Assumptions like negligible heat loss to the container or environment simplify calculations but may not always reflect real-world conditions.

• Errors can occur from external heat transfer.
• Precision instruments are critical to reliable experiments.
• Repeatability and documentation are vital for validating results.

This particular exercise teaches how to account for stray forms of energy loss and consider their impact on experimental outcomes. The importance of conducting such rigorous experiments ensures that theoretical predictions hold when applied to practical situations, advancing our technological and scientific capabilities.