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The unit shows what the numerical value represents, such as a distance in metres or a time in seconds."},{"text":"It shows the accuracy of the measurement.","correct":false,"explanation":"Units do not indicate how accurate a measurement is; accuracy depends on the method and equipment used. A unit is required because it shows the quantity the number represents."},{"text":"It shows the precision of the measurement.","correct":false,"explanation":"The number of significant figures communicates precision, not the unit. A final answer needs a unit to show which physical quantity and measurement scale the number represents."},{"text":"It shows the instrument used for measurement.","correct":false,"explanation":"The measuring instrument is identified by the experimental method, not by the unit. The unit must be included because it shows the quantity represented by the numerical value."}]},{"id":"40caea13-38fb-4084-82db-ebe503faf51c","cardType":"basic","points":2,"front":"Why is kg/m³ a derived unit?","back":"It combines kg with m³.","mcqOptions":[{"text":"It combines kg with m³.","correct":true,"explanation":"A derived unit is formed by combining base units. Kilograms per cubic metre combines kilograms for mass with cubic metres for volume, producing the unit of density."},{"text":"It uses only the base unit kg.","correct":false,"explanation":"Kilograms are a base unit, but cubic metres are derived from length. The complete unit is derived because it combines kg with m³ rather than using kg alone."},{"text":"It contains the prefix kilo.","correct":false,"explanation":"The prefix kilo is part of the SI base unit kilogram and does not make kg/m³ derived. It is derived because kilograms and cubic metres are combined."},{"text":"It divides metres by kilograms.","correct":false,"explanation":"Writing a unit with a division sign does not by itself explain its origin. kg/m³ is derived because it combines the mass unit kg with the volume unit m³."}]},{"id":"d9e37d77-45ba-436b-8885-9a9a7e788b42","cardType":"cloze","points":null,"front":"Length uses [ ]; mass uses [ ].","back":"Length uses metres (m); mass uses kilograms (kg).","mcqOptions":[{"text":"Length uses metres (m); mass uses kilograms (kg).","correct":true,"explanation":"The SI base unit of length is the metre, symbol m, while the SI base unit of mass is the kilogram, symbol kg."},{"text":"Length uses centimetres (cm); mass uses kilograms (kg).","correct":false,"explanation":"Centimetres can measure length, but the standard SI base unit is metres. Mass correctly uses kilograms, so the completed statement should use metres and kilograms."},{"text":"Length uses metres (m); mass uses grams (g).","correct":false,"explanation":"Grams measure mass, but the SI base unit of mass is the kilogram. Length correctly uses metres, so the completed statement should use metres and kilograms."},{"text":"Length uses seconds (s); mass uses newtons (N).","correct":false,"explanation":"Seconds measure time rather than length, while newtons measure force rather than mass. The correct SI base units are metres for length and kilograms for mass."}]},{"id":"bc5e98c4-a5d7-4e9d-b329-2bef73717640","cardType":"basic","points":2,"front":"Why does area use m² rather than m?","back":"Area is two lengths multiplied.","mcqOptions":[{"text":"Area is two lengths multiplied.","correct":true,"explanation":"Area is found by multiplying one length by another. Each length contributes a factor of metres, so the resulting unit is m × m = m²."},{"text":"Area is two lengths added.","correct":false,"explanation":"Adding two lengths produces another length measured in metres, not an area. Area requires multiplication of two perpendicular lengths, giving the squared unit m²."},{"text":"Area is three lengths multiplied.","correct":false,"explanation":"Three multiplied lengths describe volume and produce m³. Area has only two dimensions, so it is calculated from two lengths multiplied and uses m²."},{"text":"Area is two lengths divided.","correct":false,"explanation":"Dividing one length by another produces a dimensionless ratio because the metre units cancel. Area instead comes from multiplying two lengths, giving m²."}]},{"id":"1351d55d-5c03-4501-bfd6-16ff4f70e743","cardType":"basic","points":2,"front":"Why does volume use m³ rather than m?","back":"Volume is three lengths multiplied.","mcqOptions":[{"text":"Volume is three lengths multiplied.","correct":true,"explanation":"Volume measures three-dimensional space, so it is found by multiplying three lengths. The units multiply as m × m × m, giving m³."},{"text":"Volume is two lengths multiplied.","correct":false,"explanation":"Multiplying two lengths gives an area measured in m², not a volume. Volume includes a third dimension, so three lengths must be multiplied to give m³."},{"text":"Volume is three lengths added.","correct":false,"explanation":"Adding three lengths still produces a length measured in metres. Volume is calculated by multiplying three dimensions, which produces the cubic unit m³."},{"text":"Volume is four lengths multiplied.","correct":false,"explanation":"Multiplying four lengths would produce m⁴, which is not the unit of ordinary three-dimensional volume. Volume requires exactly three multiplied lengths and therefore uses m³."}]},{"id":"638d929c-e014-46a1-8f69-ce7d0fd0c58f","cardType":"cloze","points":null,"front":"Since 1 cm = 0.01 m, 1 cm² = [ ].","back":"Since 1 cm = 0.01 m, 1 cm² = 0.0001 m².","mcqOptions":[{"text":"0.0001 m²","correct":true,"explanation":"Area conversion factors must be squared. Squaring 0.01 gives 0.0001, so one square centimetre equals 0.0001 square metres."},{"text":"0.01 m²","correct":false,"explanation":"This applies the linear conversion factor without squaring it. Because area has two dimensions, $(0.01)^2 = 0.0001$, so the correct value is 0.0001 m²."},{"text":"0.001 m²","correct":false,"explanation":"This incorrectly treats squaring 0.01 as adding only one extra decimal place. The conversion factor must be multiplied by itself, giving 0.0001 m²."},{"text":"0.000001 m²","correct":false,"explanation":"This cubes the conversion factor, which is appropriate for volume rather than area. Squaring 0.01 gives 0.0001, so one square centimetre is 0.0001 m²."}]},{"id":"6ae3cd58-2700-4df2-b84a-1f4590b9bdbb","cardType":"cloze","points":null,"front":"Since 1 cm = 0.01 m, 1 cm³ = [ ].","back":"Since 1 cm = 0.01 m, 1 cm³ = 0.000001 m³.","mcqOptions":[{"text":"0.000001 m³","correct":true,"explanation":"Volume conversion factors must be cubed. Cubing 0.01 gives 0.000001, so one cubic centimetre equals 0.000001 cubic metres."},{"text":"0.01 m³","correct":false,"explanation":"This uses the linear conversion factor without cubing it. Volume has three dimensions, so $(0.01)^3 = 0.000001$, making the correct value 0.000001 m³."},{"text":"0.0001 m³","correct":false,"explanation":"This squares the centimetre-to-metre conversion and therefore gives an area conversion. Cubic units require cubing 0.01, which gives 0.000001 m³."},{"text":"0.001 m³","correct":false,"explanation":"This treats the cubic conversion as a factor of one thousand, often confused with litre conversions. Cubing 0.01 produces 0.000001 m³."}]},{"id":"e0d14577-c36b-4581-a834-b1e73ebb632d","cardType":"basic","points":1,"front":"Since weight is a force, what unit is weight measured in?","back":"newtons (N).","mcqOptions":[{"text":"newtons (N).","correct":true,"explanation":"Weight is the gravitational force acting on an object. All forces are measured in newtons, with the symbol N, so weight is measured in newtons."},{"text":"joules (J).","correct":false,"explanation":"Joules measure energy or work, not force. Because weight is the gravitational force on an object, its correct unit is newtons, symbol N."},{"text":"kilograms (kg).","correct":false,"explanation":"Kilograms measure mass, which is the amount of matter in an object. Weight is a force caused by gravity, so it is measured in newtons, symbol N."},{"text":"pascals (Pa).","correct":false,"explanation":"Pascals measure pressure, defined as force per unit area. Weight is a force itself, so its correct unit is newtons, symbol N."}]},{"id":"4acb7b6d-de95-4173-8c3d-e669c516356c","cardType":"cloze","points":null,"front":"Density compares mass with [ ], giving unit [ ].","back":"Density compares mass with volume, giving unit kg/m³.","mcqOptions":[{"text":"Density compares mass with volume, giving unit kg/m³.","correct":true,"explanation":"Density is mass divided by volume, so its SI unit is kilograms per cubic metre. The volume must be cubed in metres when calculating density in kg/m³."},{"text":"Density compares mass with area, giving unit kg/m².","correct":false,"explanation":"Area is two-dimensional and produces square-metre units, but density compares mass with three-dimensional volume. The correct unit is kg/m³ because volume is measured in cubic metres."},{"text":"Density compares mass with length, giving unit kg/m.","correct":false,"explanation":"Using length alone ignores the object’s three-dimensional space and gives kg/m rather than a density unit. Density is mass divided by volume, measured in kg/m³."},{"text":"Density compares mass with volume, giving unit g/cm³.","correct":false,"explanation":"Volume is the correct quantity, but this is not the requested SI unit. For density in standard SI units, mass is in kilograms and volume is in cubic metres, giving kg/m³."}]},{"id":"75eeafb8-a46f-4cd2-ab18-b228c595aec0","cardType":"basic","points":1,"front":"What equation links density, mass and volume?","back":"$\\rho = \\frac{m}{V}$.","mcqOptions":[{"text":"$\\rho = \\frac{m}{V}$.","correct":true,"explanation":"Density is defined as mass divided by volume, so the equation is ρ = m/V. Rearranging it also gives mass as density times volume and volume as mass divided by density."},{"text":"$\\rho = \\frac{V}{m}$.","correct":false,"explanation":"This reverses mass and volume, giving the reciprocal of density, sometimes called specific volume. The correct relationship places mass in the numerator: ρ = m/V."},{"text":"$m = \\frac{\\rho}{V}$.","correct":false,"explanation":"This rearrangement divides density by volume, rather than multiplying density by volume to find mass. The defining equation is ρ = m/V."},{"text":"$V = \\frac{m}{\\rho}$.","correct":false,"explanation":"This is a correct rearrangement for finding volume, but it is not the equation directly linking the three quantities in the requested density form. The answer is ρ = m/V."}]},{"id":"4b02712f-e715-40ee-8e04-56efbfae70b8","cardType":"basic","points":2,"front":"When density is needed in kg/m³ but side length is in cm, what should be converted first?","back":"Convert the side length to metres before cubing.","mcqOptions":[{"text":"Convert the side length to metres before cubing.","correct":true,"explanation":"A side length in centimetres must be converted to metres before cubing, because cubing the converted length then produces volume in m³, matching the kg/m³ density unit."},{"text":"Convert the side length to millimetres before cubing.","correct":false,"explanation":"Millimetres do not produce cubic metres after cubing, so the resulting volume would not match kg/m³ without another conversion. The side length should first be converted to metres."},{"text":"Convert the volume to metres before cubing.","correct":false,"explanation":"Volume is measured in cubic metres, not metres, and it is obtained by cubing the side length. Convert the side length to metres before carrying out the cubing."},{"text":"Convert the mass to grams before cubing.","correct":false,"explanation":"Changing mass to grams moves it away from the kilogram unit required for kg/m³ and does not solve the length-unit problem. Convert the side length to metres first."}]},{"id":"41feb53e-ecec-4089-86c8-f005ac12bfeb","cardType":"cloze","points":null,"front":"Average speed = distance ÷ time, so its unit is [ ].","back":"Average speed = distance ÷ time, so its unit is m/s.","mcqOptions":[{"text":"m/s","correct":true,"explanation":"Speed is distance divided by time. Measuring distance in metres and time in seconds gives the unit metres per second, written m/s."},{"text":"m/s²","correct":false,"explanation":"The squared seconds belong to acceleration, which is the change in velocity per second. Average speed is distance divided by time, so its unit is m/s."},{"text":"m²/s","correct":false,"explanation":"Squaring the metres would describe an area divided by time, not a distance divided by time. Speed uses metres per second, written m/s."},{"text":"s/m","correct":false,"explanation":"Reversing the division gives time per distance, related to pace rather than speed. Since speed equals distance divided by time, the unit is m/s."}]},{"id":"be5fd911-2710-4447-9c86-7632bdc1987a","cardType":"basic","points":1,"front":"What unit is acceleration measured in?","back":"m/s².","mcqOptions":[{"text":"m/s².","correct":true,"explanation":"Acceleration measures how quickly velocity changes. A velocity change in metres per second occurring each second gives metres per second squared, m/s²."},{"text":"m/s.","correct":false,"explanation":"m/s is the unit of velocity or speed, not the rate at which velocity changes. Acceleration includes an additional division by time, giving m/s²."},{"text":"N/kg.","correct":false,"explanation":"N/kg is dimensionally equivalent to m/s² in some force calculations, but the standard unit requested for acceleration here is m/s². It directly expresses velocity change per second."},{"text":"m²/s.","correct":false,"explanation":"m²/s represents area divided by time and does not describe a change in velocity. Acceleration is measured in m/s² because velocity changes by metres per second each second."}]},{"id":"4189c6dc-b541-44c4-a5b8-514fa149f16d","cardType":"basic","points":2,"front":"What does an acceleration of 2.0 m/s² mean?","back":"Velocity changes by 2.0 m/s each second.","mcqOptions":[{"text":"Velocity changes by 2.0 m/s each second.","correct":true,"explanation":"An acceleration of 2.0 m/s² means the velocity changes by 2.0 m/s during every second, assuming the acceleration remains constant."},{"text":"Velocity changes by 2.0 m/s each metre.","correct":false,"explanation":"The second in m/s² means the velocity changes per second, not per metre. Therefore, the velocity changes by 2.0 m/s each second."},{"text":"Velocity changes by 2.0 m each second.","correct":false,"explanation":"Acceleration describes a change in velocity, which includes direction and has units of m/s. Omitting the per-second part of velocity gives the wrong interpretation."},{"text":"Velocity changes by 4.0 m/s each second.","correct":false,"explanation":"The numerical value of acceleration is already 2.0, so each second adds or subtracts 2.0 m/s from the velocity. Doubling it is an arithmetic misconception."}]},{"id":"eb5c283b-64de-4d60-87e3-bc07226a973d","cardType":"basic","points":1,"front":"What equation links acceleration, velocity change and time?","back":"$a = \\frac{v-u}{t}$.","mcqOptions":[{"text":"$a = \\frac{v-u}{t}$.","correct":true,"explanation":"Acceleration equals the change in velocity divided by the time taken. Subtracting initial velocity u from final velocity v gives the velocity change, then dividing by t gives a."},{"text":"$a = \\frac{u-v}{t}$.","correct":false,"explanation":"This reverses the velocity subtraction and changes the sign of the acceleration. The change is final velocity minus initial velocity, so a = (v-u)/t."},{"text":"$a = \\frac{v+u}{t}$.","correct":false,"explanation":"Adding the velocities does not find their change and can give a non-zero result even when velocity is unchanged. Acceleration uses v-u divided by time."},{"text":"$a = \\frac{t}{v-u}$.","correct":false,"explanation":"This takes the reciprocal of the required relationship, producing time per velocity change rather than velocity change per time. Acceleration is a = (v-u)/t."}]},{"id":"a94ad559-9bd0-4607-b149-3aa399edc3e1","cardType":"basic","points":2,"front":"What does 1 N mean for a 1 kg mass?","back":"It gives 1 kg an acceleration of 1 m/s².","mcqOptions":[{"text":"It gives 1 kg an acceleration of 1 m/s².","correct":true,"explanation":"From F = ma, a force of 1 N acting on a 1 kg mass produces a = 1/1 = 1 m/s²."},{"text":"It gives 1 kg an acceleration of 1 m/s.","correct":false,"explanation":"m/s is velocity, not acceleration. Dividing force by mass gives acceleration, whose unit is m/s², so the 1 kg mass accelerates at 1 m/s²."},{"text":"It gives 1 kg an acceleration of 2 m/s².","correct":false,"explanation":"For a fixed 1 kg mass, F = ma gives 1 = 1 × a, so the acceleration is 1 m/s². Doubling the acceleration would require a 2 N force."},{"text":"It gives 2 kg an acceleration of 1 m/s².","correct":false,"explanation":"The definition specifies a 1 kg mass. A 1 N force would give a 2 kg mass only 0.5 m/s², so changing the mass makes this statement wrong."}]},{"id":"826eca17-a823-4805-b385-76015cc29b1e","cardType":"basic","points":1,"front":"Work done = force × distance, so 1 J equals what compound unit?","back":"1 N m.","mcqOptions":[{"text":"1 N m.","correct":true,"explanation":"A joule is defined as the work done when a force of one newton moves an object through one metre, so its compound unit is newton metre, N m."},{"text":"1 N/m.","correct":false,"explanation":"Dividing newtons by metres gives force per distance, not work done. Work is force multiplied by distance, so the correct compound unit is 1 N m."},{"text":"1 kg m/s.","correct":false,"explanation":"This combines mass, distance and speed rather than force and distance. Since work done equals force times distance, the correct compound unit is 1 N m."},{"text":"1 N m².","correct":false,"explanation":"The extra square on the distance gives an area factor, which is not present in the work equation. Force multiplied by distance gives 1 N m."}]},{"id":"0b8418d6-3bf3-4b69-a941-cd844a4335c3","cardType":"basic","points":1,"front":"If a calculation is force ÷ area, what unit should the answer have?","back":"pascals (Pa).","mcqOptions":[{"text":"pascals (Pa).","correct":true,"explanation":"Pressure is calculated as force divided by area, and the SI unit for pressure is the pascal, abbreviated Pa."},{"text":"newtons (N).","correct":false,"explanation":"Newtons measure force alone, before division by area. Force divided by area gives pressure, whose unit is pascals (Pa)."},{"text":"joules (J).","correct":false,"explanation":"Joules measure energy or work, which involves force multiplied by distance. Dividing force by area instead produces pressure in pascals (Pa)."},{"text":"watts (W).","correct":false,"explanation":"Watts measure power, the rate of energy transfer, not force per unit area. The unit for force divided by area is pascals (Pa)."}]},{"id":"4ea62ebd-7e83-4941-9967-d599f201947f","cardType":"cloze","points":null,"front":"One pascal equals [ ].","back":"One pascal equals 1 N/m².","mcqOptions":[{"text":"1 N/m²","correct":true,"explanation":"A pascal is defined as one newton of force acting over one square metre, so $1\\ \\text{Pa} = 1\\ \\text{N/m}^2$."},{"text":"1 N/m","correct":false,"explanation":"This uses distance rather than area in the denominator. Pressure requires force divided by square metres, so one pascal equals 1 N/m²."},{"text":"1 N m²","correct":false,"explanation":"Multiplying by square metres reverses the pressure relationship. A pascal is force per unit area, so the correct expression is 1 N/m²."},{"text":"1 kg/m²","correct":false,"explanation":"This represents mass per unit area, not force per unit area. Pressure is measured in newtons per square metre, so one pascal equals 1 N/m²."}]},{"id":"61410ab8-0672-44e6-ae57-22de8a38526e","cardType":"basic","points":1,"front":"How do you convert °C to K for IGCSE calculations?","back":"Add 273.","mcqOptions":[{"text":"Add 273.","correct":true,"explanation":"For IGCSE calculations, convert Celsius to kelvin by adding 273 to the Celsius temperature, giving the corresponding absolute temperature."},{"text":"Subtract 273.","correct":false,"explanation":"Subtracting 273 converts kelvin to approximately Celsius, not Celsius to kelvin. To convert °C to K, add 273."},{"text":"Add 100.","correct":false,"explanation":"The 100-degree interval between Celsius benchmarks does not represent the offset between the scales. Celsius to kelvin conversion requires adding 273."},{"text":"Multiply by 273.","correct":false,"explanation":"Kelvin and Celsius have the same size temperature intervals, so conversion uses an offset rather than multiplication. Add 273 to the Celsius value."}]},{"id":"5e13433b-9536-45f2-b6b1-edf4e7043bce","cardType":"basic","points":1,"front":"A temperature rises by 20°C. What is the rise in kelvin?","back":"20 K.","mcqOptions":[{"text":"20 K.","correct":true,"explanation":"A change of one degree Celsius has the same size as a change of one kelvin. Therefore, a rise of 20°C is a rise of 20 K."},{"text":"293 K.","correct":false,"explanation":"Adding 273 converts an absolute Celsius temperature to kelvin, but temperature changes do not need this offset. A rise of 20°C is 20 K."},{"text":"253 K.","correct":false,"explanation":"Subtracting 20 from 273 treats the temperature change as an absolute temperature and reverses the required reasoning. Equal Celsius and kelvin changes give 20 K."},{"text":"0.073 K.","correct":false,"explanation":"Dividing by 273 is an inappropriate conversion for a temperature interval. Celsius and kelvin intervals are equal in size, so the rise is 20 K."}]},{"id":"333a83f2-bf86-4469-805b-a7d6e5f1a603","cardType":"cloze","points":null,"front":"The kelvin scale is the [ ]; 0 K is [ ].","back":"The kelvin scale is the absolute temperature scale; 0 K is absolute zero.","mcqOptions":[{"text":"The kelvin scale is the absolute temperature scale; 0 K is absolute zero.","correct":true,"explanation":"Kelvin measures absolute temperature from a true zero point. At 0 K, particles have their minimum possible thermal energy, a point called absolute zero."},{"text":"The kelvin scale is the relative temperature scale; 0 K is the freezing point of water.","correct":false,"explanation":"Kelvin is an absolute scale, and its zero is not water’s freezing point. The correct terms are absolute temperature scale and absolute zero."},{"text":"The kelvin scale is the absolute temperature scale; 0 K is the boiling point of water.","correct":false,"explanation":"Although kelvin is an absolute scale, 0 K is far below ordinary water temperatures. Its zero point is called absolute zero, not the boiling point."},{"text":"The kelvin scale is the Celsius temperature scale; 0 K is 0°C.","correct":false,"explanation":"Kelvin and Celsius have equal-sized intervals but different zero points. Kelvin is the absolute temperature scale, and 0 K is absolute zero."}]},{"id":"29a9e1cc-c03f-4d53-8a8d-3f4f68456124","cardType":"basic","points":1,"front":"Which quantity has the unit J/kg °C?","back":"Specific heat capacity.","mcqOptions":[{"text":"Specific heat capacity.","correct":true,"explanation":"Specific heat capacity is the energy needed to raise the temperature of 1 kg of a substance by 1°C, so its unit is J/kg °C."},{"text":"Specific latent heat.","correct":false,"explanation":"Specific latent heat has units J/kg because it concerns a change of state without a temperature change. The unit J/kg °C belongs to specific heat capacity."},{"text":"Thermal conductivity.","correct":false,"explanation":"Thermal conductivity describes how readily heat passes through a material and has units involving watts, metres and kelvin. J/kg °C measures specific heat capacity."},{"text":"Heat capacity.","correct":false,"explanation":"Heat capacity describes the energy needed to raise an entire object by 1°C and is measured in J/°C. Including kg in the denominator identifies specific heat capacity."}]},{"id":"fcf66f79-07b3-4dc6-b576-5185bc04efe5","cardType":"basic","points":1,"front":"Which equation gives change in thermal energy using specific heat capacity?","back":"$\\Delta Q = m \\times c \\times \\Delta \\theta$.","mcqOptions":[{"text":"$\\Delta Q = m \\times c \\times \\Delta \\theta$.","correct":true,"explanation":"The change in thermal energy equals mass multiplied by specific heat capacity multiplied by the temperature change, so the equation is $\\Delta Q = m \\times c \\times \\Delta \\theta$."},{"text":"$\\Delta Q = m \\times c \\div \\Delta \\theta$.","correct":false,"explanation":"Dividing by the temperature change gives the wrong relationship and units. Thermal energy change increases with temperature change, so use $\\Delta Q = m \\times c \\times \\Delta \\theta$."},{"text":"$\\Delta Q = m \\div c \\times \\Delta \\theta$.","correct":false,"explanation":"Specific heat capacity measures energy required per kilogram per degree, so it must be multiplied by mass and temperature change. The correct equation is $\\Delta Q = m \\times c \\times \\Delta \\theta$."},{"text":"$\\Delta Q = c \\div m \\times \\Delta \\theta$.","correct":false,"explanation":"Dividing by mass reverses the mass relationship and produces incorrect units. Thermal energy change is proportional to mass, so $\\Delta Q = m \\times c \\times \\Delta \\theta$."}]},{"id":"7edd43d5-6b78-4ab2-af58-af4b3207f96a","cardType":"basic","points":2,"front":"What is the specific heat capacity if 0.50 kg gains 8400 J and rises by 20°C?","back":"840 J/kg °C.","mcqOptions":[{"text":"840 J/kg °C.","correct":true,"explanation":"Specific heat capacity is calculated using $c = \\frac{E}{m\\Delta T}$. Substituting the values gives $c = \\frac{8400}{0.50 \\times 20} = 840\\ \\text{J/kg °C}$, so this value is correct."},{"text":"16800 J/kg °C.","correct":false,"explanation":"Dividing the energy by the mass only gives 16,800 J/kg and ignores the temperature rise. The temperature change must also be included, giving the correct value of 840 J/kg °C."},{"text":"420 J/kg °C.","correct":false,"explanation":"Using 1 kg instead of the actual mass of 0.50 kg doubles the denominator and produces a value that is too small. The correct calculation gives 840 J/kg °C."},{"text":"8400 J/kg °C.","correct":false,"explanation":"Dividing the energy by the temperature rise but forgetting to divide by the mass gives 420 J/°C, not the specific heat capacity. Including 0.50 kg gives 840 J/kg °C."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Units","title":"Units Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"c6b41551-b15f-5798-97d0-867fba0e7b7b","walled":"$L38"}],"$L39","$L3a"]