FOCUSED REVIEW
Focused Review — Heat, Specific Heat, and Calorimetry — Calculus-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: Calculus-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
From Differential to Finite
Work the prompt, then compare with the revealed answer.
Integrate dQ = mc dT for constant m and c.
Reveal Answers
Q = ∫mc dT = mc(Tf−Ti).
Why it works: The algebraic relation is the constant-c special case of the integral model.
ACTIVITY 2
Slope Meaning
Work the prompt, then compare with the revealed answer.
What does dQ/dT represent for a fixed-mass sample?
Reveal Answers
dQ/dT = mc(T) = C(T), the sample’s heat capacity.
Why it works: It measures energy required per unit temperature rise at that temperature.
ACTIVITY 3
Variable c Setup
Work the prompt, then compare with the revealed answer.
For c(T)=a+bT, write Q from Ti to Tf.
Reveal Answers
Q = m[a(Tf−Ti) + (b/2)(Tf²−Ti²)].
Why it works: Integrate c(T) before evaluating the limits.
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
Differential Heat Capacity
For an infinitesimal temperature change, dQ = mc(T)dT. If c is constant over the interval, integration gives Q = mcΔT. If c varies with temperature, retain c(T) inside the integral.
Q = m∫c(T)dT
Example: Q = m∫ from Ti to Tf c(T)dT; constant c gives mc(Tf−Ti).
Sensei note: Do not pull c outside the integral unless treating it as constant over the interval.
KEY CONCEPT 2
Specific Heat and Heat Capacity
Specific heat c is heat capacity per unit mass, while total heat capacity is C(T)=mc(T). The slope dQ/dT equals C when the path involves only sensible heating.
dQ/dT = C(T) = mc(T)
Example: A larger C means more energy is required for the same differential temperature rise.
Sensei note: Distinguish the material property c from the object property C.
KEY CONCEPT 3
Calorimetry as an Integral Energy Balance
For an isolated calorimeter, conservation of energy requires ΣQi = 0. Each object’s heat-transfer term may be written as m∫c(T)dT between that object’s initial temperature and the common final temperature.
ΣQi = 0
Example: With constant specific heats, the integral form reduces to the familiar algebraic calorimetry equation.
Sensei note: The sign is automatic when the integration limits follow each object from Ti to Tf.
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 0.200 kg sample has c(T)=400+0.5T J/(kg·K), with T in °C for this empirical model. Find Q from 20 °C to 60 °C.
Reveal Answers
Q = 0.200∫₂₀⁶⁰(400+0.5T)dT = 3360 J.
Why it works: Because c varies with T, using a single endpoint value would not be exact.
PRACTICE 2
Guided Problem
Work the prompt, then compare with the revealed answer.
Two bodies with constant heat capacities C₁ and C₂ start at T₁ and T₂ in an isolated calorimeter. Derive Tf.
Reveal Answers
C₁(Tf−T₁)+C₂(Tf−T₂)=0, so Tf=(C₁T₁+C₂T₂)/(C₁+C₂).
Why it works: The result is a heat-capacity-weighted average of the initial temperatures.
PRACTICE 3
Independent Problem
Work the prompt, then compare with the revealed answer.
A body has heat capacity C(T)=A+BT. Write the heat absorbed from Ti to Tf and identify the constant-C limit.
Reveal Answers
Q=A(Tf−Ti)+(B/2)(Tf²−Ti²). If B=0, Q=AΔT.
Why it works: The finite energy transfer is the area under the C-versus-T curve.
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
Integral Setup
Work the prompt, then compare with the revealed answer.
When must Q = m∫c(T)dT be used instead of mcΔT?
Reveal Answers
When c varies significantly with temperature over the interval and a variable-c model is supplied or required.
Why it works: Constant c is an approximation, not a mathematical identity for all materials and ranges.
QUICK CHECK 2
Weighted Equilibrium
Work the prompt, then compare with the revealed answer.
If C₁ = 2C₂, which initial temperature has twice the weight in Tf?
Reveal Answers
T₁, because Tf=(C₁T₁+C₂T₂)/(C₁+C₂).
Why it works: Larger heat capacity makes the equilibrium temperature stay closer to that object’s initial temperature.
QUICK CHECK 3
Energy Conservation
Work the prompt, then compare with the revealed answer.
Write the isolated two-body balance in integral form.
Reveal Answers
m₁∫ from T₁ to Tf c₁(T)dT + m₂∫ from T₂ to Tf c₂(T)dT = 0.
Why it works: Using each object’s own limits automatically captures heating versus cooling signs.
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
General relation
Use dQ = mc(T)dT and integrate when specific heat varies with temperature.
KEY TAKEAWAY 2
Constant-c limit
For constant c, the integral reduces to Q = mcΔT.
KEY TAKEAWAY 3
Calorimetry conservation
For an isolated system, sum each object’s finite heat transfer and set the total equal to zero.
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?
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