Required Practical: Investigating Specific Heat Capacity

Purpose of the Practical

This practical is designed to measure the specific heat capacity of a material. Specific heat capacity is the amount of energy needed to raise the temperature of 1 kilogram of a substance by 16C. By investigating this, you will understand how energy is transferred to a material when it is heated and how this affects its temperature.

The experiment involves heating a block of material and measuring the temperature change as energy is supplied. This helps you link the electrical energy supplied to the block with the thermal energy gained by the block.

Equipment and Setup

• Insulated block of the material to be tested (e.g., aluminium or copper)
• Electric heater (immersion heater or heating coil) connected to a power supply
• Thermometer or temperature sensor to measure temperature changes accurately
• Ammeter to measure current flowing through the heater
• Voltmeter to measure potential difference across the heater
• Stopwatch or timer to record heating time
• Balance to measure the mass of the block

The block should be well insulated to reduce heat loss to the surroundings, for example by using a foam jacket or wrapping it in insulating material.

Method

1. Measure the mass of the block using a balance and record it.
2. Insert the heater into the block and connect it to the power supply.
3. Place the thermometer or temperature sensor in the block to measure its temperature.
4. Record the initial temperature of the block before heating.
5. Turn on the power supply and start the stopwatch to begin heating the block.
6. Record the current (I) from the ammeter and the potential difference (V) from the voltmeter.
7. After a set time (t), turn off the power supply and record the final temperature of the block.
8. Calculate the temperature change $\Delta T = T_{\text{final}} - T_{\text{initial}}$.

By supplying electrical energy to the heater, energy is transferred to the block, raising its temperature. Measuring the current, voltage, and time allows you to calculate the energy supplied.

Calculations and Analysis

The energy transferred to the block by the heater is calculated using the formula:

$$E = V \times I \times t$$

where:

• $E$ is the energy transferred in joules (J)
• $V$ is the potential difference in volts (V)
• $I$ is the current in amperes (A)
• $t$ is the time in seconds (s)

Once you have the energy supplied and the temperature change, you can calculate the specific heat capacity $c$ of the material using:

$$c = \frac{E}{m \times \Delta T}$$

where:

• $c$ is the specific heat capacity in joules per kilogram per degree Celsius (J/kg6C)
• $m$ is the mass of the block in kilograms (kg)
• $\Delta T$ is the temperature change in degrees Celsius (6C)

It is important to consider sources of error and how they affect the accuracy of your results:

• Heat loss to the surroundings despite insulation will cause energy to be lost, making the calculated specific heat capacity higher than the true value.
• Inaccurate temperature readings due to slow response or poor placement of the thermometer.
• Assuming all electrical energy supplied is transferred to the block; in reality, some energy is lost as sound or heat to the surroundings.
• Measurement errors in current, voltage, time, or mass.

To improve accuracy:

• Use better insulation to reduce heat loss.
• Ensure the thermometer is well placed for accurate temperature measurement.
• Repeat the experiment and calculate an average value.
• Use digital sensors for more precise measurements.

Learning example:

A 2.0 kg aluminium block is heated using a heater connected to a 12 V power supply. The current through the heater is 3.0 A. The block
9s temperature rises from 206C to 306C in 200 seconds. Calculate the specific heat capacity of aluminium.

First, calculate the energy supplied:

$$E = V \times I \times t = 12 \times 3.0 \times 200 = 7200 \text{ J}$$

Calculate the temperature change:

$$\Delta T = 30 - 20 = 10^\circ \text{C}$$

Calculate specific heat capacity:

$$c = \frac{E}{m \times \Delta T} = \frac{7200}{2.0 \times 10} = \frac{7200}{20} = 360 \text{ J/kg}^\circ \text{C}$$

This value is lower than the actual specific heat capacity of aluminium (~900 J/kg6C), suggesting heat loss or measurement errors. Experimental values often differ from theoretical values due to such factors.

Note: This experimental value is lower than the accepted specific heat capacity of copper (~385 J/kg6C) due to heat loss and measurement inaccuracies.

Note: This experimental value is lower than the accepted specific heat capacity of iron (~450 J/kg6C) due to heat loss and measurement inaccuracies.

Diagram suggestion: Add a labelled diagram showing the experimental setup including the insulated block, heater, power supply, ammeter, voltmeter, thermometer, and insulation to aid understanding.