Subject: Physics | Level: GCSE | Exam Board: OCR
This guide provides a comprehensive, exam-focused breakdown of Specific Heat Capacity (SHC) for OCR GCSE Physics (6.4). It covers the core concepts, the essential equation you must memorise, the required practical (PAG 1), and common pitfalls to help you secure every possible mark on this frequently tested topic.
Revision Notes & Key Concepts
Revision Podcast Transcript
Hello and welcome to Physics Unlocked — the podcast that turns your GCSE revision into something you actually want to listen to. I'm your tutor for today, and in this episode we're diving deep into one of the most calculation-heavy topics in OCR Gateway Physics A: Specific Heat Capacity. This comes up in Topic P1, Matter, and it is absolutely guaranteed to appear on your exam — so let's make sure you're completely ready for it. Whether you're sitting Foundation or Higher tier, this episode covers everything you need. We'll go through the core concepts, nail the equation, walk through the required practical, and then I'll give you my top exam tips and a quick-fire quiz at the end. Grab a pen and paper — you're going to want to take notes. --- SECTION ONE: CORE CONCEPTS — What is Specific Heat Capacity? Let's start with the big question: what actually is specific heat capacity? Imagine you're heating two different substances — a block of aluminium and a beaker of water — using the same heater for the same amount of time. You'd notice that the aluminium heats up much faster than the water. Why? Because water requires far more energy to raise its temperature by the same amount. That resistance to temperature change is what we call specific heat capacity. Here's the formal definition, and I want you to write this down because examiners will ask you to state it: Specific heat capacity is the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius. The unit is joules per kilogram per degree Celsius — written as J stroke kg stroke degrees C, or sometimes J per kg per K if you're using Kelvin, though for GCSE we almost always use degrees Celsius. Now, the key equation. This is the most important thing in this entire topic, and you must memorise it because it is not given to you on the OCR formula sheet. Ready? Here it is: Delta E equals m times c times delta T. Let's break that down. Delta E is the change in thermal energy, measured in joules. m is the mass of the substance, measured in kilograms. c is the specific heat capacity, in joules per kilogram per degree Celsius. And delta T is the change in temperature, in degrees Celsius. So: change in thermal energy equals mass times specific heat capacity times temperature change. Now, here's a really important analogy to help this stick. Think of specific heat capacity like a sponge. A substance with a HIGH specific heat capacity is like a big, thick sponge — it soaks up a lot of energy before its temperature rises much. Water has a specific heat capacity of 4200 joules per kilogram per degree Celsius — that's enormous. Aluminium, by contrast, is about 900. So water is like a giant sponge; aluminium is like a thin cloth. Same energy input, very different temperature rise. This is why water is used as a coolant in car engines and in central heating systems — it can absorb huge amounts of thermal energy without getting dangerously hot. --- SECTION TWO: REARRANGING THE EQUATION Let's talk about rearranging the equation, because OCR examiners love to give you a question where you need to find c, or find the mass, rather than just the energy. The equation is: Delta E = m × c × ΔT To find c: divide both sides by m and delta T. So c = Delta E divided by (m times delta T). To find m: m = Delta E divided by (c times delta T). To find delta T: delta T = Delta E divided by (m times c). I want to give you a worked example right now. Listen carefully. A 2 kilogram block of aluminium is heated from 20 degrees Celsius to 65 degrees Celsius. The specific heat capacity of aluminium is 900 joules per kilogram per degree Celsius. Calculate the change in thermal energy stored by the block. Step one: identify your values. Mass m = 2 kg. Specific heat capacity c = 900 J per kg per degree C. Temperature change delta T = 65 minus 20 = 45 degrees Celsius. Step two: substitute into the equation. Delta E = 2 times 900 times 45. Step three: calculate. 2 times 900 is 1800. 1800 times 45 is 81,000. Final answer: Delta E = 81,000 joules, or 81 kilojoules. Notice I showed every step. In a 4-mark calculation, OCR will award marks for: writing the correct equation, correct substitution, correct calculation, and correct units. Miss the units and you lose a mark. Always write J for joules. --- SECTION THREE: THE REQUIRED PRACTICAL — PAG 1 Now let's talk about the required practical, because this is tested heavily in OCR exams. This is Practical Activity Group 1, or PAG 1, and it involves measuring the specific heat capacity of a solid, typically an aluminium block. Here's the apparatus: you need an aluminium block with two holes drilled into it — one for an immersion heater and one for a thermometer. You connect the heater to a power supply and either a joulemeter, or a voltmeter and ammeter so you can calculate the energy. You wrap the block in cotton wool to reduce energy loss to the surroundings. The method: record the starting temperature. Switch on the heater. Record the temperature every minute for around ten minutes. Record the energy supplied using the joulemeter, or calculate it using Energy = Power times time, where Power = Voltage times Current. Then rearrange the SHC equation to find c: c = Energy divided by (mass times temperature change). Now here's what the examiners really want you to know about this practical. First: why do we insulate the block? The answer is NOT just "to keep it warm" or "to reduce heat loss." You must say: to reduce the transfer of energy to the surroundings, so that more of the electrical energy supplied goes into heating the block rather than the surrounding air. This gets you the mark. Second: what is the effect of NOT insulating? If energy is lost to the surroundings, the temperature rise of the block will be smaller than expected. When you calculate c using the equation, you divide the energy by the mass and the temperature change. If delta T is smaller than it should be, you're dividing by a smaller number, which gives you a LARGER value of c. So the calculated specific heat capacity will be higher than the true value. This is a systematic error. Examiners love asking this — remember: energy lost to surroundings means your calculated c is too high. Third: why do we use a joulemeter rather than just timing? Because a joulemeter directly measures the electrical energy supplied to the heater, which is more accurate than calculating it from voltage, current, and time — especially if the current fluctuates. --- SECTION FOUR: EXAM TIPS AND COMMON MISTAKES Right, this is the section that could be worth five or six marks on its own. These are the mistakes I see candidates make time and time again, and I want you to avoid every single one of them. Mistake number one: not converting grams to kilograms. If the question gives you a mass of 500 grams, you MUST divide by 1000 to get 0.5 kilograms before substituting into the equation. The unit of mass in the SHC equation is always kilograms. I cannot stress this enough — this is the single most common error in SHC calculations. Mistake number two: not converting kilojoules to joules. If the question says a heater supplies 4.5 kilojoules of energy, you must multiply by 1000 to get 4500 joules. The unit of energy in the equation is always joules. Mistake number three: confusing specific heat capacity with specific latent heat. These are two completely different things. Specific heat capacity applies when a substance is changing temperature — the substance stays in the same state. Specific latent heat applies when a substance is changing state — melting, freezing, boiling, condensing — and crucially, the temperature does NOT change during a state change. If a question mentions "melting" or "boiling" or "change of state," that's specific latent heat, not specific heat capacity. Mistake number four: vague answers about energy loss. Never write "heat is lost." Always write "energy is transferred to the surroundings" or "energy is lost to the atmosphere." Examiners will not credit vague language. Mistake number five: forgetting to calculate delta T. If the question says the temperature rises from 18 degrees to 63 degrees, you must calculate delta T = 63 minus 18 = 45 degrees. Do not substitute 63 into the equation — that would be the final temperature, not the change. Now, a tip for 4 to 6 mark questions that involve power. If you're given the power of a heater in watts and the time it runs for in seconds, you first need to calculate the total energy using E = P times t. Then use that energy value in the SHC equation. This is a two-step calculation and OCR often uses it to test whether candidates can link equations across topics. For command words: if the question says "Calculate," always show your working, write the equation first, substitute values, and give units in your final answer. If it says "Explain," use the word "because" to link cause and effect — for example, "The calculated value of c is higher than the true value because energy is transferred to the surroundings, reducing the temperature rise of the block." If it says "Describe," say what happens — no need to explain why unless asked. --- SECTION FIVE: QUICK-FIRE RECALL QUIZ Right, it's quiz time! I'll ask a question, give you five seconds to think, then give the answer. Ready? Question one: What is the equation for change in thermal energy? ... The answer is: Delta E equals m times c times delta T. Question two: What are the units of specific heat capacity? ... The answer is: joules per kilogram per degree Celsius — J stroke kg stroke degrees C. Question three: A student calculates a specific heat capacity that is higher than the accepted value. What does this suggest about the experiment? ... The answer is: energy was lost to the surroundings during the experiment. Question four: What must you do to a mass given in grams before using it in the SHC equation? ... Divide by 1000 to convert to kilograms. Question five: What is the specific heat capacity of water? ... 4200 joules per kilogram per degree Celsius. --- SECTION SIX: SUMMARY AND SIGN-OFF Let's bring it all together. Here are the five things you absolutely must know for your OCR exam. One: The equation is Delta E = m times c times delta T. Memorise it — it's not on the formula sheet. Two: Units matter. Mass in kilograms, energy in joules, temperature change in degrees Celsius, specific heat capacity in J per kg per degrees C. Three: In the PAG 1 practical, insulation reduces energy transfer to the surroundings. Without it, your calculated c will be too high. Four: Specific heat capacity is about temperature change. Specific latent heat is about state change. Don't confuse them. Five: In multi-step calculations, use E = P times t first to find the energy, then substitute into the SHC equation. That's everything for today's episode of Physics Unlocked. You've covered the core concept, the equation and its rearrangements, the required practical, common exam mistakes, and you've tested yourself with a quick-fire quiz. You are now significantly better prepared for your OCR exam than you were twenty minutes ago — and that's what this is all about. Good luck with your revision, and remember: every mark you practise for now is a mark you'll earn on exam day. See you in the next episode!
Key Terms & Definitions
- Specific Heat Capacity (c)
- The amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius.
- Change in Thermal Energy (ΔE)
- The total amount of energy transferred to or from a substance to change its temperature. Measured in Joules (J).
- Temperature Change (ΔT)
- The difference between the final temperature and the initial temperature of a substance (T_final - T_initial).
- Systematic Error
- An error that is consistent and repeatable, often caused by the experimental apparatus or method, which causes readings to be skewed in one direction.
- Insulation
- A material that reduces the rate of thermal energy transfer.
- Joulemeter
- A device that measures the amount of electrical energy (in Joules) supplied to a component.
Worked Examples
Worked Example
Question: A 0.5 kg block of copper is heated by a 50 W heater for 2 minutes. The temperature of the block increases from 20 °C to 51 °C. Calculate the specific heat capacity of the copper. [6 marks]
Solution: **Step 1: Calculate the total energy supplied (ΔE).** The question gives power (P) and time (t), so we must first use E = P × t. Time must be in seconds: 2 minutes = 2 × 60 = 120 s. P = 50 W, t = 120 s ΔE = 50 W × 120 s = 6000 J **Step 2: Calculate the temperature change (ΔT).** ΔT = Final Temperature - Initial Temperature ΔT = 51 °C - 20 °C = 31 °C **Step 3: State the SHC equation and rearrange it for 'c'.** ΔE = m × c × ΔT c = ΔE / (m × ΔT) **Step 4: Substitute the known values into the rearranged equation.** m = 0.5 kg, ΔE = 6000 J, ΔT = 31 °C c = 6000 / (0.5 × 31) **Step 5: Calculate the final answer.** c = 6000 / 15.5 c = 387.096... J/kg°C **Step 6: State the final answer to an appropriate number of significant figures, with units.** Final answer: **c = 390 J/kg°C** (to 2 s.f.)
Worked Example
Question: A student performs an experiment to find the specific heat capacity of water. Her calculated value is 4500 J/kg°C. The accepted value is 4200 J/kg°C. Explain why the student's value is higher than the accepted value. [3 marks]
Solution: Step 1: Identify the primary source of error. Energy was transferred from the apparatus to the surroundings. Step 2: Link this error to the measurements taken. This means the measured temperature change (ΔT) of the water was less than it should have been, as not all the supplied energy went into heating the water. Step 3: Explain the effect on the final calculation. Because c = ΔE / (m × ΔT), dividing by a smaller value for ΔT results in a calculated value for 'c' that is larger than the true value.
Worked Example
Question: Calculate the energy needed to heat 750 g of iron from 15 °C to 250 °C. The specific heat capacity of iron is 450 J/kg°C. [4 marks]
Solution: **Step 1: Convert mass from grams to kilograms.** m = 750 g / 1000 = 0.75 kg **Step 2: Calculate the temperature change (ΔT).** ΔT = 250 °C - 15 °C = 235 °C **Step 3: State the SHC equation and substitute values.** ΔE = m × c × ΔT ΔE = 0.75 kg × 450 J/kg°C × 235 °C **Step 4: Calculate the final answer with units.** ΔE = 79312.5 J Final answer: **79,000 J** or **79 kJ** (to 2 s.f.)
Practice Questions
Question: State the definition of specific heat capacity. [2 marks]
Answer:
Question: A 200 g sample of oil is heated from 10 °C to 40 °C using 11.4 kJ of energy. Calculate the specific heat capacity of the oil. [5 marks]
Answer:
Question: Describe an experiment to determine the specific heat capacity of a block of aluminium. [6 marks]
Answer:
Question: A kettle has a power rating of 2.2 kW. It is used to heat 1.5 kg of water from 15 °C to its boiling point of 100 °C. Calculate the minimum time this should take. The specific heat capacity of water is 4200 J/kg°C. [5 marks]
Answer:
Question: Explain why the actual time taken to boil the water in the kettle from the previous question would be longer than the calculated minimum time. [2 marks]
Answer:
