Specific heat capacity

1. Overview

Specific heat capacity explains why different materials require different amounts of energy to change their temperature. Understanding this property is essential for everything from engineering cooling systems in car engines to predicting weather patterns near the ocean.

Key Definitions

• Internal Energy: The total energy stored by the particles that make up a system (the sum of their kinetic and potential energies).
• Temperature: A measure of the average kinetic energy of the particles in a substance.
• Specific Heat Capacity ($c$): The energy required per unit mass per unit temperature increase.

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Core Content

Temperature and Internal Energy

• When you heat an object, you are transferring energy to its internal energy store.
• A rise in the temperature of an object is a direct result of an increase in its internal energy.
• If the internal energy increases, the particles within the object move faster (in a gas or liquid) or vibrate more vigorously (in a solid).

Worked Example (Core Concepts): Question: Two identical 1kg iron blocks are heated. Block A reaches 50°C and Block B reaches 100°C. Which has more internal energy? Answer: Block B has more internal energy because it has a higher temperature, indicating more energy has been transferred to its particles.

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Extended Content (Extended Only)

Particles and Kinetic Energy

• At a microscopic level, an increase in temperature is defined as an increase in the average kinetic energy of all the particles in the object.
• The faster the particles move/vibrate, the higher the temperature recorded by a thermometer.

Defining Specific Heat Capacity Every substance has a unique Specific Heat Capacity ($c$). It tells us how many Joules of energy are needed to heat 1kg of that substance by 1°C.

• Water has a very high SHC (~4200 J/kg°C), meaning it takes a lot of energy to heat up and cools down slowly.
• Metals usually have low SHC, meaning they heat up and cool down very quickly.

Experimental Measurement of Specific Heat Capacity To find the SHC of a substance (e.g., a metal block or a liquid), you must measure the mass, the energy supplied, and the temperature change.

1. Experiment for a Solid (Metal Block):

• Measure the mass ($m$) of the block using a balance.
• Place an immersion heater in one hole and a thermometer in the other.
• Use a power source and a Joulemeter to measure the energy ($E$) supplied.
• Record the initial and final temperature to find the change ($\Delta \theta$).
• Insulation: Wrap the block in cotton wool to prevent heat loss to the surroundings, which would make the calculated SHC value too high.

2. Experiment for a Liquid:

• Measure the mass of the empty beaker, then the mass of the beaker with liquid to find the mass ($m$) of the liquid.
• Use a heater, thermometer, and Joulemeter as above.
• Stirring: Use a stirrer to ensure the temperature is uniform throughout the liquid before taking a reading.

Worked Example (Extended): A 0.5 kg copper block is heated from 20°C to 60°C. If 7,800 J of energy was supplied, calculate the Specific Heat Capacity of copper.

1. Identify the variables: $m = 0.5$ kg, $\Delta E = 7800$ J, $\Delta \theta = (60 - 20) = 40$ °C.
2. Use the formula: $c = \Delta E / (m \times \Delta \theta)$
3. Calculate: $c = 7800 / (0.5 \times 40) = 7800 / 20 = 390$ J/kg°C.

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Key Equations

The Specific Heat Capacity Equation: $$\Delta E = m c \Delta \theta$$ OR $$c = \frac{\Delta E}{m \Delta \theta}$$

• $\Delta E$: Change in thermal energy (Joules, J)
• $m$: Mass (kilograms, kg)
• $c$: Specific Heat Capacity (J/kg°C)
• $\Delta \theta$: Change in temperature (°C)

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Common Mistakes to Avoid

• ❌ Wrong: Thinking the Specific Heat Capacity of a substance changes if you add more heat.
  • ✓ Right: SHC is a constant property of a material. Adding more energy just results in a higher temperature rise, not a different SHC.
• ❌ Wrong: Forgetting to convert grams to kilograms.
  • ✓ Right: Mass must be in kg for the standard formula. If a question gives 500g, you must use 0.5kg in your calculation.
• ❌ Wrong: Choosing a substance like Mercury as a "good heat store" because it gets hot quickly.
  • ✓ Right: Water is a better heat store because it has a high SHC, meaning it can hold a lot of energy without a massive rise in temperature.
• ❌ Wrong: Calculating the temperature change and forgetting to add it to the starting temperature.
  • ✓ Right: If a question asks for the final temperature, calculate $\Delta \theta$ first, then add it to the initial temperature.

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Exam Tips

1. Check the units: If the energy is given in kJ, multiply by 1,000 to get Joules before starting your calculation.
2. Rearrange carefully: Use a formula triangle if you struggle with algebra to ensure you don't divide the wrong numbers.
3. Experimental Errors: If an exam question asks why your experimental value for SHC is higher than the textbook value, the answer is almost always "Heat was lost to the surroundings."

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Exam-Style Questions

Practice these original exam-style questions to test your understanding. Each question mirrors the style, structure, and mark allocation of real Cambridge 0625 Theory papers.

Exam-Style Question 1 — Short Answer [5 marks]

Question:

A student heats a 500 g block of copper using an electric heater. The heater supplies energy at a rate of 100 W. Assume all the energy from the heater is transferred to the copper block.

(a) State the formula that relates the energy supplied, the mass of the copper, the specific heat capacity of the copper, and the temperature change. [1]

(b) Define specific heat capacity. [2]

(c) The temperature of the copper block increases from 20°C to 40°C in a time of 6 minutes. Calculate the specific heat capacity of the copper. [2]

Worked Solution:

(a)

1. $E = mc\Delta\theta$ [States the correct formula]

How to earn full marks:

• Must have all terms correct: energy (E), mass (m), specific heat capacity (c), and temperature change (Δθ).

(b)

1. Specific heat capacity is the energy required to raise the temperature of 1 kg of a substance by 1 °C (or 1 K). [Defines specific heat capacity with all key elements]

How to earn full marks:

• 1 mark for mentioning energy required.
• 1 mark for mentioning per unit mass and per unit temperature change.

(c)

1. Calculate the energy supplied: $E = Pt = 100 \times (6 \times 60) = 36000 , J$ [Calculates the energy supplied in Joules]
2. Rearrange the formula and substitute to find specific heat capacity: $c = \frac{E}{m\Delta\theta} = \frac{36000}{0.5 \times (40-20)} = 3600 , J/(kg \cdot {}^\circ C)$ [Correctly rearranges the formula and substitutes values]
3. $c = 3600 , J/(kg \cdot {}^\circ C)$ [States the final answer with correct units]

How to earn full marks:

• 1 mark for calculating the energy supplied in Joules.
• 1 mark for correct substitution and final answer with units. $\boxed{3600 , J/(kg \cdot {}^\circ C)}$

Common Pitfall: Remember to convert the mass into kilograms before using it in the specific heat capacity formula. Also, always include the correct units in your final answer; losing the units will cost you a mark.

Exam-Style Question 2 — Extended Response [8 marks]

Question:

A student performs an experiment to determine the specific heat capacity of water. They use a 1.0 kW immersion heater to heat 0.50 kg of water in a well-insulated container. The initial temperature of the water is 20°C. The heater is switched on for 120 seconds.

(a) Calculate the energy supplied by the heater to the water. [2]

(b) The student measures the final temperature of the water to be 70°C. Calculate the specific heat capacity of water using the student’s data. [3]

(c) Suggest two reasons why the value of the specific heat capacity calculated in (b) might be different from the accepted value of 4200 J/(kg °C). [2]

(d) Suggest one improvement to the experiment that would give a more accurate value for the specific heat capacity of water. [1]

Worked Solution:

(a)

1. Calculate the energy supplied: $E = Pt = 1000 \times 120 = 120000 , J$ [Calculates the energy supplied in Joules]

How to earn full marks:

• 1 mark for using the correct formula.
• 1 mark for correct substitution and answer with units. $\boxed{120000 , J}$

(b)

1. Calculate the temperature change: $\Delta\theta = 70 - 20 = 50 , {}^\circ C$ [Calculates the temperature change]
2. Rearrange the formula and substitute to find specific heat capacity: $c = \frac{E}{m\Delta\theta} = \frac{120000}{0.50 \times 50} = 4800 , J/(kg \cdot {}^\circ C)$ [Correctly rearranges the formula and substitutes values]
3. $c = 4800 , J/(kg \cdot {}^\circ C)$ [States the final answer with correct units]

How to earn full marks:

• 1 mark for calculating the temperature change.
• 1 mark for correct substitution into the specific heat capacity formula.
• 1 mark for the final answer with correct units. $\boxed{4800 , J/(kg \cdot {}^\circ C)}$

(c)

1. Heat loss to the surroundings: Some energy is lost from the container to the surroundings, meaning not all the energy supplied by the heater goes into heating the water.
2. Heater not fully immersed: The heater might not be fully immersed in the water, leading to some energy being lost to the air instead of heating the water.

How to earn full marks:

• 1 mark for each valid reason. Acceptable answers include: heat loss to the surroundings, incomplete insulation, heater not fully immersed, thermometer not accurate.

(d)

1. Use a lid on the container. [Suggests a valid improvement to reduce heat loss]

How to earn full marks:

• 1 mark for a valid improvement such as: use a lid, use more insulation, use a more accurate thermometer.

Common Pitfall: When calculating the temperature change, make sure to subtract the initial temperature from the final temperature. Also, remember that even a "well-insulated" container will still lose some heat to the surroundings, affecting the accuracy of the experiment.

Exam-Style Question 3 — Short Answer [6 marks]

Question:

A student wants to investigate the specific heat capacity of aluminum. They have a 200 g aluminum block and a heater that provides a known amount of energy.

(a) Describe an experimental procedure the student could use to determine the specific heat capacity of the aluminum block. Include the measurements they would need to take. [4]

(b) State two precautions the student should take to ensure the experiment is accurate. [2]

Worked Solution:

(a)

1. Measure the initial temperature of the aluminum block using a thermometer. [States the first step: measuring initial temperature]
2. Place the heater in a hole drilled in the aluminum block and insulate the block to reduce heat loss. [States the importance of insulation]
3. Supply a known amount of energy to the block using the heater (measure voltage, current, and time if using an electrical heater). [States how to supply and measure the energy]
4. Measure the final temperature of the aluminum block after a certain time. [States the measurement of the final temperature]

How to earn full marks:

• 1 mark for measuring the initial temperature.
• 1 mark for mentioning insulation to reduce heat loss.
• 1 mark for describing how to supply and measure the energy.
• 1 mark for measuring the final temperature.

(b)

1. Ensure good thermal contact between the heater and the block by using thermal paste. [States a precaution for good thermal contact]
2. Stir the block while heating to ensure even heat distribution. [States a precaution to ensure even heat distribution]

How to earn full marks:

• 1 mark for each valid precaution. Acceptable answers include: good thermal contact, stir the block, use a sensitive thermometer, reduce draughts.

Common Pitfall: Don't forget the importance of good thermal contact between the heater and the block. Air gaps can significantly reduce the efficiency of heat transfer. Also, stirring helps to ensure a uniform temperature throughout the block.

Exam-Style Question 4 — Extended Response [9 marks]

Question:

A hot water tank contains 150 kg of water at a temperature of 60°C. The tank loses heat to the surroundings at a rate of 500 W. The specific heat capacity of water is 4200 J/(kg °C).

(a) Calculate the energy lost by the water tank in 1 hour. [2]

(b) Calculate the temperature decrease of the water in the tank after 1 hour, assuming no water is added or removed from the tank. [3]

(c) The tank is fitted with an immersion heater rated at 2.0 kW. The heater is switched on to maintain the water temperature at 60°C. Explain why the water temperature remains constant, even though the heater is supplying energy. [2]

(d) Suggest two ways to reduce the rate of heat loss from the water tank to the surroundings. [2]

Worked Solution:

(a)

1. Calculate the time in seconds: $t = 1 \times 60 \times 60 = 3600 , s$ [Calculates the time in seconds]
2. Calculate the energy lost: $E = Pt = 500 \times 3600 = 1800000 , J$ [Calculates the energy lost in Joules]

How to earn full marks:

• 1 mark for calculating the time in seconds.
• 1 mark for correct substitution and answer with units. $\boxed{1800000 , J}$

(b)

1. Rearrange the formula to find the temperature change: $\Delta\theta = \frac{E}{mc} = \frac{1800000}{150 \times 4200} = 2.857 , {}^\circ C$ [Correctly rearranges the formula and substitutes values]
2. $\Delta\theta = 2.86 , {}^\circ C$ [States the final answer with correct units, rounded to 3 s.f.]

How to earn full marks:

• 1 mark for correct rearrangement of the formula.
• 1 mark for correct substitution.
• 1 mark for the final answer with correct units. $\boxed{2.86 , {}^\circ C}$

(c)

1. The rate of energy supplied by the heater is equal to the rate of energy lost to the surroundings. [States the balance of energy rates]
2. Therefore, the temperature remains constant as the energy gained is equal to the energy lost. [Explains the constant temperature]

How to earn full marks:

• 1 mark for stating that the energy supplied by the heater equals the energy lost to the surroundings.
• 1 mark for explaining that this balance maintains a constant temperature.

(d)

1. Increase the thickness of the insulation around the tank. [Suggests a way to improve insulation]
2. Reduce the surface area of the tank exposed to the surroundings. [Suggests reducing surface area]

How to earn full marks:

• 1 mark for each valid suggestion. Acceptable answers include: increase insulation thickness, use better insulation material, reduce surface area exposed to surroundings.

Common Pitfall: Pay attention to the units! The power is given in Watts, and time should be converted to seconds to get the energy in Joules. Also, remember that a constant temperature means the rate of energy input equals the rate of energy output.