FOCUSED REVIEW
Focused Review — Heat, Specific Heat, and Calorimetry — Algebra-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Heat, Specific Heat, and Calorimetry.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
RESOURCE: Physics Sensei Unit Review | UNIT ID: THM-U03 | TOPIC: Heat, Specific Heat, and Calorimetry | COURSE LEVEL: Algebra-Based College Physics
BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 15 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Equation Recall
Work the prompt, then compare with the revealed answer.
Write the two equations that organize this unit.
Reveal Answers
Q = mcΔT and ΣQ = 0 for an insulated calorimetry system.
Why it works: The first models each object; the second connects the objects.
ACTIVITY 2
Sign Reasoning
Work the prompt, then compare with the revealed answer.
A metal cools from 90 °C to 30 °C. What sign does its Q have?
Reveal Answers
ΔT = 30 − 90 = −60 °C, so Q < 0.
Why it works: Cooling means the object releases energy.
ACTIVITY 3
One-Step Calculation
Work the prompt, then compare with the revealed answer.
How much heat raises 0.250 kg of water by 4.0 °C? Use c = 4186 J/(kg·°C).
Reveal Answers
Q = (0.250)(4186)(4.0) = 4.19×10³ J.
Why it works: Convert grams to kilograms when needed.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
Sensible Heat: Q = mcΔT
When a substance remains in one phase and c is approximately constant, the heat transfer associated with a temperature change is Q = mc(Tf − Ti). The sign is built into ΔT.
Q = mc(Tf − Ti)
Example: For 0.200 kg of water warmed 5.0 °C: Q = (0.200)(4186)(5.0) ≈ 4.19×10³ J.
Sensei note: Keep mass and specific-heat units compatible before substituting.
KEY CONCEPT 2
Specific Heat and Thermal Inertia
Specific heat c measures thermal response per unit mass. For fixed Q and m, ΔT = Q/(mc), so a larger c produces a smaller temperature change.
ΔT = Q/(mc)
Example: Equal masses receiving equal heat: the sample with half the c has twice the ΔT.
Sensei note: Do not confuse specific heat c with heat capacity C = mc.
KEY CONCEPT 3
Calorimetry: ΣQ = 0
For an insulated calorimeter with negligible heat loss, the algebraic sum of heat transfers is zero. Write one Q = mc(Tf − Ti) term for each participating object and solve for the unknown.
ΣQ = 0
Example: Mixing 0.100 kg water at 80 °C with 0.200 kg water at 20 °C gives Tf = 40 °C when heat loss is neglected.
Sensei note: Use one common final equilibrium temperature, but each object keeps its own initial temperature, mass, and c.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Worked Example
Work the prompt, then compare with the revealed answer.
A 0.300 kg aluminum sample, c = 900 J/(kg·°C), warms from 20 °C to 50 °C. Find Q.
Reveal Answers
ΔT = 30 °C. Q = (0.300)(900)(30) = 8100 J.
Why it works: Positive Q means the sample absorbed heat.
PRACTICE 2
Guided Problem
Work the prompt, then compare with the revealed answer.
Mix 0.100 kg water at 80 °C with 0.200 kg water at 20 °C in an insulated cup. Find Tf.
Reveal Answers
Same c cancels: 0.100(Tf−80)+0.200(Tf−20)=0. Thus 0.300Tf=12.0 and Tf=40 °C.
Why it works: Check that Tf lies between the two initial temperatures.
PRACTICE 3
Independent Problem
Work the prompt, then compare with the revealed answer.
A 0.050 kg metal at 100 °C is placed in 0.100 kg water at 20 °C. The final temperature is 24 °C. Find the metal specific heat. Use cwater = 4186 J/(kg·°C).
Reveal Answers
Heat gained by water = (0.100)(4186)(4)=1674.4 J. Set |Qmetal| equal to this: cmetal ≈ 441 J/(kg·°C).
Why it works: Write the full energy balance before solving for the unknown c.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Solve for Heat
Work the prompt, then compare with the revealed answer.
A 0.50 kg material with c = 800 J/(kg·°C) cools by 10 °C. Find Q.
Reveal Answers
Q = (0.50)(800)(−10) = −4000 J.
Why it works: The negative sign represents heat released.
QUICK CHECK 2
Solve for Temperature Change
Work the prompt, then compare with the revealed answer.
A 2.0 kg sample with c = 500 J/(kg·°C) absorbs 6000 J. Find ΔT.
Reveal Answers
ΔT = 6000/[(2.0)(500)] = 6.0 °C.
Why it works: Isolate ΔT algebraically before substituting.
QUICK CHECK 3
Calorimetry Check
Work the prompt, then compare with the revealed answer.
Why must an ideal two-object final temperature lie between the initial temperatures?
Reveal Answers
One object must cool while the other warms until they reach equilibrium; without another energy source, the final temperature cannot exceed both or fall below both.
Why it works: This is a fast physical reasonableness check.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Model each object
Use Q = mc(Tf − Ti) for each object that changes temperature without a phase change.
KEY TAKEAWAY 2
Conserve energy
For an insulated calorimeter, ΣQ = 0.
KEY TAKEAWAY 3
Check the result
Signs and the location of Tf between initial temperatures provide fast error checks.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
Next Step
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