shc

Quick Quiz

• Specific Heat Capacity: Refers to the energy required to increase the temperature of a substance.

  • It signifies how challenging it is to heat a material up.

Different Substances Have Different Specific Heat Capacities

1. Different substances require varying amounts of energy to increase their thermal energy stores and, therefore, their temperature.

2. Materials that require significant energy to increase their temperature also release considerable energy when they cool down.

   • Such materials can store extensive amounts of energy.

3. Definition of Specific Heat Capacity: It is the energy needed to raise the temperature of 1 kg of a substance by 1°C.

4. Energy Transfer Equation:

   • The equation that connects energy transferred to specific heat capacity is: $\Delta E = mc \Delta \theta$

     • Where:

     • $\Delta E$ is the change in thermal energy (J)

     • $m$ is the mass (kg)

     • $c$ is the specific heat capacity (J/kg°C)

     • $\Delta \theta$ is the change in temperature (°C)

Investigating Specific Heat Capacities

Practical Method

1. Obtain a solid block with two holes.

2. Measure the mass of the block and wrap it in an insulating material (like thick newspaper).

3. Insert a thermometer and heater, then connect the circuit as illustrated.

4. Measure the initial temperature; set the power supply voltage (V) to 10V. Turn on the supply and start a stopwatch.

5. Record temperature and current (I) readings every minute for 10 minutes.

   • Expect current through the circuit to remain constant as the block heats up.

6. Turn off the power supply after collecting enough data.

   • Calculate power ($P = VI$).

   • Use $E = Pt$, where $t$ is time in seconds to obtain energy transfer at each temperature reading.

   • Assume all energy supplied to the heater transfers to the block.

7. Plot a graph of temperature against energy transferred to the block's thermal energy store.

   • Find the gradient of the straight part of the graph to calculate the specific heat capacity.

Specific Heat Capacity Calculation

• The formula to relate the gradient and specific heat capacity is:
  $c = \frac{1}{\text{gradient} \times m}$

• Calculation of Final Temperature: To find the final temperature of 5 kg of water at 5°C after receiving 50 kJ of energy.

Conservation of Energy Principle

• The principle of conservation of energy states that energy is always conserved; it cannot be created or destroyed.

  • Energy can be transferred usefully, stored, or dissipated.

  • During energy transfer, not all is utilized effectively; there is always a degree of energy dissipation, often considered 'waste'.

• Closed Systems: In a closed system, energy is not lost; rather, energy is dissipated within the system, with a net change of energy equating to zero.

Examples of Energy Conservation

Mobile Phone Example

• A mobile phone operates as a system where chemical energy from the battery gets transferred.

  • Part of this energy dissipates to the thermal energy store of the phone and surroundings.

Insulated Flask Example

• Dropping a cold spoon into hot soup within a sealed flask demonstrates energy transfer from the soup.

  • The soup cools as energy is transferred to the spoon, representing energy dissipation within a closed system.

Power Definition and Calculation

1. Power: Identified as the rate at which energy is transferred or work is done.

2. Unit of Power: Measured in watts (W).

   • $1 ext{ watt} = 1 ext{ joule of energy transferred per second}$.

3. Power Calculation Formulas: $P = \frac{\text{Energy Transferred} \ (J)}{\text{Time} \ (s)}$

   • $P = \frac{\text{Work Done} \ (J)}{\text{Time} \ (s)}$

Example of Power Calculation

• Two cars with identical features but different engine powers race the same distance.

  • The car with the more powerful engine reaches the finish line faster, indicating a higher energy transfer rate.

Example Problem

• Lifting a stunt performer requires 11,000 J to reach the top of a building.

  • Motor A lifts the performer in 55 seconds and motor B in 300 seconds. The power of Motor A is calculated:

  • $P = \frac{11000 \ J}{55 \ s} = 200 \ W$

  • Motor A is more powerful due to faster energy transfer.

Summary of Energy Dissipation

• Energy cannot be created or destroyed, only converted or dissipated.

  • When energy is dissipated, it persists but is no longer stored in a useful form.

Calculation of Power Output

• A motor transfers 4.8 kJ of energy in 2 minutes (120 seconds). Calculate power output.

  • Use the formula:

  • Convert kJ to J for calculation and apply $P = \frac{E}{t}$

  • Example solution submission needed to confirm calculations.

Quick Quiz - Specific Heat Capacity: Refers to the energy needed to increase a substance's temperature.

Different Substances Have Different Specific Heat Capacities

1. Varying amounts of energy are required for different substances to change temperature, impacting energy storage.

2. Definition: Energy required to raise 1 kg of a substance by 1°C.

3. Energy Transfer Equation: $ext{ΔE} = mc ext{Δθ}$

   • Where:

     • $ ext{ΔE}$ = change in thermal energy (J)

     • $m$ = mass (kg)

     • $c$ = specific heat capacity (J/kg°C)

     • $ ext{Δθ}$ = change in temperature (°C)

Investigating Specific Heat Capacities

1. Obtain a solid block with holes and measure its mass.

2. Wrap it in insulation and insert a thermometer and heater.

3. Start heating and record temperature and current readings.

4. Calculate power ($P = VI$) and energy transfer.

5. Plot temperature against energy transferred.

6. Calculate specific heat capacity using: $c = rac{1}{ ext{gradient} imes m}$

Conservation of Energy Principle

• Energy cannot be created or destroyed, only converted or dissipated.

• In closed systems, energy is conserved but may dissipate as waste.

Power Definition and Calculation

1. Power: Rate of energy transfer or work done, measured in watts (W).

2. Formulas:

   • $P = rac{ ext{Energy Transferred} ext{(J)}}{ ext{Time} ext{(s)}}$

   • $P = rac{ ext{Work Done} ext{(J)}}{ ext{Time} ext{(s)}}$

Example Problem

• Lifting a stunt performer requires 11,000 J. Motor A lifts in 55 seconds: $P = rac{11000 ext{ J}}{55 ext{ s}} = 200 ext{ W}$

Summary of Energy Dissipation

• Energy persists after dissipation but loses usefulness.

Calculation of Power Output

• A motor transfers 4.8 kJ in 120 seconds: $P = rac{E}{t}$.o