Problem 36 - 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 \(0.780 \mathrm{kg},\) 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) Average specific heat is approximately 2505.4 J/kg°C. (b) It's an underestimate if heat loss occurs.

Step 1: Calculate the Heat Energy Supplied

To find the total heat energy supplied, multiply the power of the electrical resistor by the time for which it was active. The formula is: \[ Q = P imes t \]where \( Q \) is the heat energy in joules, \( P \) is the power in watts, and \( t \) is the time in seconds. Substituting the given values,\[ Q = 65.0 \, \text{W} \times 120 \, \text{s} = 7800 \, \text{J} \].

Step 2: Determine the Temperature Change of the Liquid

Find the difference between the final and initial temperatures of the liquid. The formula for temperature change is:\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]Substituting the given temperatures,\[ \Delta T = 22.54^\circ \text{C} - 18.55^\circ \text{C} = 3.99^\circ \text{C} \].

Step 3: Calculate the Average Specific Heat Capacity

Use the formula for specific heat capacity, which is:\[ c = \frac{Q}{m \cdot \Delta T} \]where \( c \) is the specific heat capacity, \( Q \) is the heat energy supplied, \( m \) is the mass of the liquid, and \( \Delta T \) is the temperature change. Substituting the values,\[ c = \frac{7800 \, \text{J}}{0.780 \, \text{kg} \times 3.99^\circ \text{C}} = \frac{7800}{3.1122} \approx 2505.4 \, \text{J/kg}^\circ\text{C} \].

Step 4: Explanation for Part (b)

If heat loss to the container or surroundings is not negligible, the calculated specific heat assumes all heat energy went into heating the liquid. Therefore, the true amount of heat energy that increased the liquid's temperature is less than calculated, meaning the actual specific heat would be higher. Hence, the specific heat in part (a) is an underestimate.

Key Concepts

Heat Energy Transfer

In physics, heat energy transfer refers to the movement of thermal energy from one place or substance to another. This can occur through conduction, convection, or radiation. In our exercise, heat energy is specifically transferred from an electrical resistor to a liquid.
The key factor here is the resistor's power, denoted as "P," which measures how much energy it can produce over time. From our formula \[ Q = P \times t \]we know that the total heat energy "Q" is a product of this power and the time "t" that the energy is applied. The measurements given, 65.0 W of power for 120 seconds, yield a total energy transfer of 7800 Joules.
It's essential to understand that without a proper medium, or path, such as a resistor, heat can't be transferred effectively. This setup ensures the heat energy goes into changing the state of the liquid, helping us further explore the concepts of specific heat capacity and heat loss.

Specific Heat Capacity

Specific heat capacity is a material's ability to absorb heat energy without undergoing a significant temperature change. It is a critical factor in determining how substances react when exposed to heat.
The specific heat capacity "c" can be calculated using the relationship:\[ c = \frac{Q}{m \cdot \Delta T} \]where "Q" is the heat energy supplied, "m" is the mass, and "ΔT" is the change in temperature of the substance.
In our problem, the specific heat of the mystery liquid was calculated by using the data gathered: - Heat energy: 7800 J
- Mass: 0.780 kg
- Temperature change: 3.99°C
Substituting these values into the equation gives us a result of approximately 2505.4 J/kg°C.
This specific heat value teaches us about the thermal properties of the liquid, indicating how much energy is needed to raise the temperature of one kilogram of the liquid by one degree Celsius. Substances with higher specific heat capacities absorb more heat without much change in temperature.

Heat Loss

Heat loss refers to the transfer of heat energy from an environment or system to its surroundings. Typically, this is an undesired outcome in experiments where precision is key.
The exercise requires us to consider what happens when heat loss can't be ignored. If the liquid loses heat to the container or to the environment, less energy is available to increase its temperature. This is significant because, in understanding specific heat capacity, we initially assumed that all the energy went into raising the liquid's temperature. However, some energy might have been lost to the surroundings, making our original calculations potentially inaccurate.
If heat were lost, the results we calculated for specific heat are underestimated. This is because, with real heat loss, more actual heat energy would have been needed to achieve the measured temperature change in the liquid. Recognizing and measuring heat loss can be crucial in experiments that require high accuracy and precision.