DSE Physics | Specific heat capacity and latent heat lesson plan (for tutors)

DSE Physics | Specific heat capacity and latent heat lesson plan (for tutors) | E = mcDT and melting ice

2026-11-03

Tutor lesson-prep guides · Secondary Physics

Students can quote E = mcDT but forget that temperature holds still while ice melts. This lesson fixes three numbers: 336 kJ to raise 2 kg of water by 40 K, 167 kJ to melt 0.5 kg of ice, and 252 s for a 2 kW kettle.

Objectives — Students can: (1) use E = mcDT to get 336 kJ for 2 kg of water rising 40 K; (2) use E = mL to get 167 kJ for 0.5 kg of ice and explain the constant temperature; (3) finish an ice-water mixture calculation reaching 36.7 degrees Celsius; (4) compute 252 seconds for a 2 kW kettle to heat 1.5 kg from 20 to 100 degrees, and tell conduction, convection and radiation apart.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill	Quick check	Pass mark
Conservation of energy	Can energy disappear?	No, it only changes form
Power	What is the unit and meaning of power?	Watt, one joule per second
Mass and weight	Which unit does m take in a heat calculation?	Kilograms, not newtons

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-10 min	Opener	Why must you wait for a kettle but ice cools a drink at once?	Write both observations
10-24 min	Specific heat capacity	E = mcDT with c = 4200 for 2 kg rising 40 K	336 kJ
24-38 min	Latent heat	L = 3.34e+05 J/kg for 0.5 kg of ice	167 kJ
38-50 min	Ice and water mixture	1 kg at 60 C plus 0.2 kg of ice: melt first, then warm	36.7 C
50-56 min	Three modes of transfer	2 kW kettle needs 252 s; one example each	Explain the flask
56-60 min	Exit check	Mark on the spot	Three correct

03 Teaching script (read it out)

Scene 1: Specific heat capacity is what slows a temperature change

Water at 4200 J per kg per kelvin means 1 kg needs 4200 J for a 1 K rise. So 2 kg rising 40 K needs 2 x 4200 x 40 = 336 kJ. That number explains why the sea lags behind the beach and why water cools car engines. Three places marks die: mass left in grams, mixing Celsius and kelvin in the answer even though the difference is the same, and treating c as per gram. Write the units on every line and the error exposes itself.

Ask them to: compute the energy to take 500 g of water from 25 to 75 degrees.

Scene 2: Latent heat: the temperature stops, the energy does not

A latent heat of fusion of 3.34e+05 J/kg means 1 kg of ice at 0 degrees takes 334 kJ to become water at 0 degrees, so 0.5 kg needs 167 kJ, which would warm the same mass of water by 80 K. All of that energy breaks the lattice instead of raising the reading. Vaporisation is larger still, about 2.26 x 10^6 J/kg, which is why sweating works. Rule for mixtures: first ask whether the ice can all melt; if yes continue warming, if no stop at zero.

Follow up: which stores more energy, 0 degree ice or 0 degree water, and why?

Scene 3: Each mode has a carrier; a flask blocks all three

Conduction passes energy by particle collisions and is fastest in metals, which is why pans are metal and handles are plastic. Convection carries energy with a moving fluid, which is why heaters sit low and air conditioners high. Radiation is electromagnetic and needs no medium, which is why it crosses a vacuum, and black rough surfaces absorb and emit best. A vacuum flask uses a double vacuum wall against conduction and convection, a silvered surface against radiation and a cork against conduction. Power check: 2000 W for 1.5 kg from 20 to 100 degrees needs 504 kJ, so 252 seconds; real kettles take longer, which is the door into efficiency questions.

Ask them: why must the stopper of a vacuum flask be a poor conductor?

04 Question types and worked steps

Type	Example	Steps	Answer
Calculation	Energy from E = mcDT	Kilograms and kelvin first, then multiply	2 kg by 40 K gives 336 kJ
Calculation	Energy from E = mL	Mass times latent heat, no temperature term	0.5 kg gives 167 kJ
Calculation	Final temperature of a mixture	Melt the ice, then warm the rest	36.7 C
Explanation	Separating the three modes	Ask whether a medium flows or is needed at all	Collisions, flow, waves

Sanity check (teach this): Unit check: E = 2.0 x 4200 x 40 = 336000 J, which is 336 kJ, a sensible size for a kettle of water. Mixture check: 1 kg at 60 degrees holds 252 kJ, melting 0.2 kg of ice takes 66.8 kJ, leaving 185.2 kJ for 1.2 kg, so the rise is 36.75 K and the final temperature is 36.7 degrees, between 0 and 60 as expected. Kettle check: 504 kJ divided by 2 kW is 252 s, matching everyday experience.

05 Independent practice (3 questions)

Question	Answer	What to watch
Energy to raise 2 kg of water by 40 K	336 kJ	2 x 4200 x 40
Energy to melt 0.5 kg of ice	167 kJ	0.5 x 3.34e+05
Time for a 2 kW kettle to heat 1.5 kg from 20 to 100 C	252 s	504 kJ / 2 kW

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using mcDT while melting	“Melting uses E = mcDT”	“A change of state uses E = mL; the temperature does not move.”
Leaving mass in grams	Substituting 500 with c = 4200	“c is per kilogram, so convert the mass first.”
Mixing joules and kilojoules	Answering 336 J	“2 kg rising 40 K is 336 kJ, not 336 J.”
Calling convection conduction	“The air conditioner conducts cool air round the room”	“Moving air carries heat, so it is convection; conduction needs contact.”

07 Homework (paper, section and question numbers)

Physics chapter index — All chapters
Physics Paper 1 Section A — Questions 1-4: heat and latent heat

08 Exit check (last 5 minutes)

Question	Answer	Decision
Energy to raise 2 kg of water by 40 K	336 kJ	Wrong: redo the unit conversion
What happens to temperature while ice melts	It stays constant	Wrong: return to script two
Three features of a vacuum flask	Vacuum, silvered wall, cork	Two or more wrong: add transfer examples next lesson

09 Tutor FAQ

Must c and L be memorised?

Remember 4200 and 3.34 x 10^5 for water; other values are given.

Celsius or kelvin for the difference?

The number is identical because it is a difference, but be consistent in the answer.

Why is a real kettle slower?

Heat escapes to the air and the body, so efficiency is below 100 per cent.

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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.