DSE Physics: Heat and Gases — Tutor Lesson Plan (Q = mc dT, latent heat, pV/T)

2026-10-28

Tutor lesson-prep guides · Secondary physics

Students handle Q = mc dT but drop the melting stage when ice is heated to steam. This lesson turns energy questions into a staged checklist.

Objectives — the student can compute energy in stages with Q = mc dT and Q = ml, separate specific from latent heat, and solve gas law questions in kelvin.

01 Prerequisites (3-minute check)

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

Skill	Quick check	Pass mark
Heat and temperature	How do they differ?	Temperature is mean kinetic energy
Kelvin	What is 27 degrees Celsius in kelvin?	300 K
Mass units	How many grams in 1 kg?	1000 g

02 Lesson flow (60 minutes)

Time	Stage	Tutor does	Student does
0-8 min	Temperature and internal energy	The microscopic picture	State the mean kinetic idea
8-22 min	Specific heat	Q = mc dT with 2 kg of water rising 30 K	Get 252 000 J
22-34 min	Change of state	Q = ml for 0.5 kg of ice	Get 167 000 J
34-48 min	Gas laws	pV / T constant, kelvin only	Get 2.667 dm3
48-56 min	Staged energy	Ice to water to steam, three stages	Write three lines
56-60 min	Exit check	Three questions marked	Finish alone

03 Teaching script (read it out)

Scene 1: Specific heat: let the units lead

Q = mc dT with c = 4200 J/kg/K for water. Two kilograms rising from 20 to 50 degrees gives 2 x 4200 x 30 = 252 000 J.
A temperature difference in kelvin equals one in celsius, but say which one you used.

Student explains why a longer heating time does not always mean a bigger temperature rise.

Scene 2: Change of state: temperature holds still while energy goes in

Melting uses latent heat, Q = ml with l = 334 000 J/kg. Half a kilogram needs 167 000 J and stays at 0 degrees.
The energy breaks the packing between particles instead of raising mean kinetic energy, so a thermometer shows nothing.

Note: melt then heat is two stages, and the second starts at 0 degrees.

Scene 3: Gas laws: convert to kelvin before substituting

pV / T is constant. Two dm3 at 300 K heated to 400 K at fixed pressure gives 2 x 400 / 300 = 2.667 dm3.
Changing pressure too: 3 dm3 at 1 atm and 300 K to 2 atm and 450 K gives 1 x 3 x 450 / (2 x 300) = 2.25 dm3.

Student calculates the volume change from 300 K to 600 K at fixed pressure (double).

04 Question types and worked steps

Type	Example	Steps	Answer
Specific heat	Q = mc dT	2 x 4200 x 30	252 000 J
Latent heat	Q = ml	0.5 x 334 000	167 000 J
Gas law	pV / T constant	2 x 400 / 300	2.667 dm3
Mixing	hot and cold water	Heat lost equals heat gained	Solve for final T

Sanity check (teach this): A final mixture temperature must sit between the two starting temperatures. A result like 120 degrees means the equation or a mass is wrong.

05 Independent practice (3 questions)

Question	Answer	What to watch
2 kg of water rising 30 K	252 000 J	Multiplied by the change, not the final value
0.5 kg of ice melted at 0 degrees	167 000 J	Second stage added
2 dm3 at 300 K to 400 K, fixed pressure	2.667 dm3	Kelvin used first

06 Common mistakes and what to say

Mistake	Student writes	Say this
Using final temperature for dT	energy doubled	dT is the rise, not the reading.
Only one stage in a melting question	warming stage missing	After melting it is 0 degrees, then it warms.
Celsius in a gas law	volume ratio wrong	Gas laws need kelvin.
Latent and specific heat swapped	mc dT used for melting	If temperature does not change, use ml.

07 Homework (paper, section and question numbers)

DSE Physics practice paper (heating and energy) — Section B (heating 0.5 kg of water)
Secondary physics chapter index — All chapters

08 Exit check (last 5 minutes)

Question	Answer	Decision
Energy for 2 kg of water rising 30 K	252 000 J	All three right: move to gas laws
Energy to melt 0.5 kg of ice	167 000 J	One wrong: re-teach latent heat
2 dm3 at 300 K to 400 K	2.667 dm3	Two wrong: repeat scene 3

09 Tutor FAQ

Can dT be in celsius?

Yes, the difference is the same, but a gas law needs kelvin.

Why does melting not raise temperature?

Energy goes into breaking the arrangement between particles.

How many constants to memorise?

Water c = 4200 and ice l = 334 000; others are given.

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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.