Calorimetry and Thermal Equilibrium
Calorimetry applies energy conservation to thermally interacting bodies: when external heat and work are negligible, the algebraic sum of all sensible and latent heat changes from the initial states to equilibrium is zero.
Why this shows up in the exam
Mixing liquids at different temperatures · Measuring unknown specific heat · Including vessels and calorimeters in heat balance
Learn the idea
In an isolated calorimeter, total heat lost equals total heat gained at the common final temperature. Imagine energy being transferred internally until every interacting part reaches one temperature; the algebraic heat changes, including the vessel, must sum to zero.
🧠 Memory hook: One final temperature, one signed energy ledger.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- sum_i Q_i = 0 — isolated-system calorimetry with heat gains positive and losses negative
- Q_i = m_i c_i(T_f-T_i) — sensible heat of component i without phase change
- C_cal = m_water-equivalent c_water — calorimeter heat capacity represented by water equivalent
How to approach it
- 1Propose a physically possible final state
- 2Write every component's path to that state
- 3Set net heat to zero and verify the assumed state
Common slip-ups that cost marks
- •Averaging temperatures without heat-capacity weights
- •Ignoring the calorimeter
- •Solving for a final state outside the assumed phase range
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A wire 1 m long and cross-sectional area 2 mm^2 extends by 1 mm under a 200 N load. Find Young modulus.
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