Problem 42 - University Physics with Modern 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 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 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 2558 J/kg°C. (b) It's an overestimate due to potential heat loss.

Step 1: Calculate the Total Heat Added

The electrical energy supplied is converted to heat. Given the power, \( P = 65.0 \, \text{W} \), and the time duration, \( t = 120 \, \text{s} \), the total energy (heat) supplied is calculated using the equation: \[ Q = P \times t \]. So, \( Q = 65.0 \, \text{W} \times 120 \, \text{s} = 7800 \, \text{J} \).

Step 2: Calculate the Temperature Change

The initial temperature of the liquid is \( 18.55^{\circ} \text{C} \) and the final temperature is \( 22.54^{\circ} \text{C} \). The temperature change is given by: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} = 22.54^{\circ} \text{C} - 18.55^{\circ} \text{C} = 3.99^{\circ} \text{C} \].

Step 3: Calculate the Specific Heat

The formula for specific heat \( c \) is \[ c = \frac{Q}{m \Delta T} \], where \( Q \) is the heat added, \( m \) is the mass of the liquid, and \( \Delta T \) is the temperature change. Substituting the known values: \[ c = \frac{7800 \, \text{J}}{0.780 \, \text{kg} \times 3.99^{\circ} \text{C}} \approx 2558 \, \text{J/kg} \cdot \text{C} \].

Step 4: Consider Heat Lost to Surroundings

If some heat is transferred to the container or surroundings, the specific heat calculated in step 3 would be an overestimate. This is because not all the electrical energy is used to increase the temperature of the liquid; some is lost, meaning the actual energy used to heat the liquid is less than calculated.

Key Concepts

Thermal Energy Transfer

Thermal energy transfer occurs when energy moves from one system to another, usually in the form of heat. In experimental physics, measuring this transfer is crucial to understand how much energy is involved in processes like heating liquids. In the exercise, the technician used an electrical resistor to convert electrical energy into heat, then transferred this heat to the liquid.

Understanding thermal energy transfer helps comprehend how substances change temperature and how much energy is required for these changes.

• Total heat added to a system can be computed using the formula \( Q = P \times t \), where \( P \) is the power output in watts and \( t \) is the time in seconds.
• In this context, \( 7800 \, \text{J} \) was the energy transferred to the liquid during the experiment, as calculated by multiplying the power, \( 65.0 \, \text{W} \), by the time span, \( 120 \, \text{s} \).
• No heat is lost to the surroundings in an ideal situation, focusing all calculated energy on changing the liquid's temperature.

By focusing on thermal energy transfer, scientists determine specific heat capacities and other properties, vital for understanding material behavior.

Experimental Physics

Experimental physics involves conducting controlled experiments to understand physical phenomena, often by measuring quantities like energy or temperature. In our exercise, experimental physics was showcased by determining the specific heat of a liquid through measurements and controlled interventions.

The process in the experiment:

• An electrical resistor immersed in the liquid allows scientists to introduce a known quantity of energy into the system.
• It's essential to monitor the temperature change accurately, as the temperature shift directly informs about the energy interaction with the liquid.
• Observations and results from such experiments, like the rise from \( 18.55^{\circ} \mathrm{C} \) to \( 22.54^{\circ} \mathrm{C} \), guide in understanding thermal processes.

This careful regulation and observation ensure the reliability of results, leading to valid conclusions about physical properties. Experimental physics offers a platform for testing theoretical predictions and contributes to developing practical applications.

Heat Transfer Calculations

Heat transfer calculations are pivotal for deducing crucial properties, like specific heat capacity, of materials. These calculations help us quantify how much heat is needed to achieve a certain temperature change in a substance.

Here's how the calculation works in our scenario:

• The specific heat capacity \( c \) can be calculated using the formula \( c = \frac{Q}{m \Delta T} \), where \( Q \) is the thermal energy supplied, \( m \) is the mass, and \( \Delta T \) is the temperature change.
• In the exercise, \( c = \frac{7800 \, \text{J}}{0.780 \, \text{kg} \times 3.99^{\circ} \mathrm{C}} \approx 2558 \, \text{J/kg} \cdot \text{C} \), representing the specific heat capacity of the liquid.
• Such calculations help in understanding how efficient a material is at storing thermal energy.

Furthermore, in practical experiments, possibilities of heat exchange with the environment or containers often require consideration. If not considered, it can lead to inaccuracies, as seen when not accounting for heat lost, which may lead to overestimates of specific heat values. Calculating heat transfer vividly illustrates the energy needs and stability of materials regarding temperature variations.