Problem 32 - 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 s at a constant rate of 65.0 W. The mass of the liquid is 0.780 kg, and its temperature increases from 18.55\(^\circ\)C to 22.54\(^\circ\)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) Average specific heat is approximately 2560 J/(kg·°C). (b) The result is an overestimate due to neglecting heat loss.

Step 1: Define the known quantities

First, let's list the known values from the problem:- Power of the resistor: \( P = 65.0 \, \text{W} \)- Time: \( t = 120 \, \text{s} \)- Mass of the liquid: \( m = 0.780 \, \text{kg} \)- Initial temperature: \( T_i = 18.55 \, ^\circ\text{C} \)- Final temperature: \( T_f = 22.54 \, ^\circ\text{C} \)

Step 2: Calculate the heat energy transferred to the liquid

The energy transferred to the liquid (Q) can be calculated as the product of power and time. \[ Q = P \times t \]Substituting the known values gives:\[ Q = 65.0 \, \text{W} \times 120 \, \text{s} = 7800 \, \text{J} \]

Step 3: Calculate the temperature change in the liquid

The temperature change (\( \Delta T \)) of the liquid can be found by subtracting the initial temperature from the final temperature. \[ \Delta T = T_f - T_i = 22.54 \, ^\circ\text{C} - 18.55 \, ^\circ\text{C} = 3.99 \, ^\circ\text{C} \]

Step 4: Find the average specific heat capacity

The specific heat (c) can be found using the formula for heat transfer:\[ Q = m \times c \times \Delta T \]Rearrange to solve for c:\[ c = \frac{Q}{m \times \Delta T} \]Substitute the known values:\[ c = \frac{7800 \, \text{J}}{0.780 \, \text{kg} \times 3.99 \, ^\circ\text{C}} \approx 2560 \, \text{J/(kg} \cdot ^\circ\text{C)} \]

Step 5: Analyze the impact of heat loss

If heat transfer from the liquid to the container or surroundings is considered, the energy used to increase the temperature of the liquid would be reduced by the amount of heat lost. Thus, the calculated specific heat in part (a) would be an overestimate since the denominator in our calculation is too large; we're assuming all 7800 J went to heating the liquid, when some energy was actually lost.

Key Concepts

Heat Transfer

Heat transfer is a key concept in understanding specific heat capacity. It involves the movement of thermal energy from one substance to another. In our exercise, the electrical resistor transfers heat to the liquid, causing its temperature to rise.
Heat transfer is governed by the principle of conservation of energy, which means the energy put into the system is accounted for as heat gained or lost. In our case, the power of 65.0 W is converted into 7800 J of heat energy over a time of 120 seconds.
Understanding this exchange helps in calculating the specific heat capacity. The heat gained by a liquid is expressed using the formula:

• \( Q = m \times c \times \Delta T \)
• Where \( Q \) is the heat transferred, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

In this setup, we assume no heat is lost to the container or surroundings to simplify calculations.

Energy Conversion

Energy conversion is crucial in studying how electrical energy is transformed into thermal energy. In experimental setups like this, a resistor is used to convert electrical energy into heat.
Let's break down the process:

• The electrical energy enters the resistor as a power of 65.0 W.
• This energy is sustained for 120 seconds, providing a total of 7800 J.
• The resistor converts most of that energy into heat, which then warms up the liquid.

By understanding energy conversion, we're able to appreciate how different forms of energy can be utilized and measured during scientific experiments.

Thermal Physics

Thermal physics combines principles from both classical mechanics and thermodynamics to explain how systems exchange heat.
The specific heat capacity, a vital part of thermal physics, tells us how much heat is required to raise the temperature of a given mass by 1°C. We see this concept when determining the specific heat capacity of the unknown liquid.
Consider the relevant formula of specific heat capacity:

• \( c = \frac{Q}{m \times \Delta T} \)
• Where \( Q \) is the heat added, \( m \) is the mass, and \( \Delta T \) is the change in temperature.

This relationship helps us understand the intrinsic properties of different materials based on their response to heat application. Thermal physics provides insight into the nature of heat and temperature, impacting various scientific and engineering applications.

Experimental Methods

Experimental methods are techniques used to systematically gather data and find results in a scientific investigation.
When determining specific heat capacity, it's key to ensure accurate measurements and controlled conditions. Here are some essential points:

• Ensure precise measurement of mass, temperature change, and time to guarantee the correct calculation of heat transfer.
• Minimize heat loss to the surroundings to avoid errors in the measurement of thermal properties.
• It's important to assume minimal influences from external factors, like the container absorbing any heat, which might require an adjustment and verification of assumptions.

These methods help in producing reliable results and are fundamental to experimental physics. Practical adjustments and observations, like recognizing potential heat loss, improve the reliability of experimental outcomes.