Specific heat capacity — calorimetry: Visual walkthrough

1.7.4 · D2 · Physics › Thermodynamics › Specific heat capacity — calorimetry

Hum is order mein build karte hain: "heat" aur "temperature" pictures ke roop mein kya hain → heat, mass aur temperature change par kyun depend karti hai → single-body law $Q = m c Δ T$ → sealed-cup bookkeeping → meeting point ke liye solve karna → degenerate cases. Har symbol (jinmein $1$ aur $2$ subscripts bhi shamil hain jo do bodies ko label karte hain) pehli baar aate hi define ho jaata hai — kuch bhi pehle use nahi hota.

Step 1 — Temperature matlab "atoms kitna zyada hil rahe hain"

KYA HAI. Koi bhi formula aane se pehle, ek solid ko chhote balls (atoms) ki grid ke roop mein socho jo springs se jude hain. Ye kabhi bhi still nahi hote — ye vibrate karte hain. Temperature ek measure hai ki ye kitni zyada violence se, average mein, hilte hain. Jitna zyada garam = utna bada hilna.

YE YAHAN SE KYUN SHURU KAREIN. Baad mein aane wale har symbol — $Q$ , $c$ , $Δ T$ — secretly is hilne ke baare mein hi hain. Agar hum picture nahi karte ki "warm up hona" aslam mein kya hota hai, toh formula sirf letters hain. Same idea words mein Heat and Internal Energy note mein dekho.

PICTURE. Left side mein, cold atoms thoda sa hilte hain (chhote arrows). Right side mein, wahi atoms bahut zyada hilte hain (lambe arrows). Sirf shake ka size badla, kuch nahi badla.

Figure — Specific heat capacity — calorimetry

Step 2 — Heat $Q$ woh energy hai jo tum IN daalo taaki jiggling bade

KYA HAI. Atoms ko zyada hilane ke liye, tumhe unhe energy deni padegi. Woh transferred energy heat kehlati hai, likha jaata hai $Q$ , measure hoti hai joules (J) mein. Heat daalne se $T$ badhti hai; heat nikalne se ghatti hai.

NAYA SYMBOL KYUN. Temperature ek state hai (abhi kitni jiggly hai). Heat ek flow hai (energy andar ya bahar cross kar rahi hai). Humein do alag symbols chahiye kyunki ye do alag sawaalon ke jawaab dete hain: "kitna garam?" versus "kitni energy move hui?". Yahi distinction First Law of Thermodynamics ka poora kamal hai.

PICTURE. Ek flame energy ke arrows ( $Q$ ) ball-and-spring grid mein daalta hai; arrows bade jiggles ke roop mein land hote hain. Sign convention: andar ki taraf arrows ⇒ $Q > 0$ ; bahar ki taraf arrows ⇒ $Q < 0$ .

Step 3 — Zaruri heat mass $m$ ke proportional kyun hoti hai

KYA HAI. Ek block lo, phir do identical blocks lo. Dono ko same amount warm karne ke liye, tumhe do gune zyada atoms ko jiggle karna hoga — toh do guni heat chahiye.

MULTIPLY KYUN, ADD KYUN NAHI. Har atom ko zyada jiggle karne ke liye apna share of energy chahiye. Double atoms (double mass $m$ ) ⇒ double total energy. Yeh ek proportionality hai: $Q \propto m$ . Hum " $\propto$ " likhte hain (padho "is proportional to") yeh kehne ke liye ki "kisi constant ke through lockstep mein badhta hai".

PICTURE. Ek block heat ka ek bucket soakta hai $Δ T$ badhne ke liye; do blocks stacked ko same rise ke liye do buckets chahiye.

Step 4 — Yeh temperature rise $Δ T$ ke bhi proportional kyun hai

KYA HAI. Ab mass fix karo aur zyada badi rise maango. 20 K warm karne ke liye 10 K warm karne se do guni heat chahiye.

KYUN. Jiggle ki har extra degree ki cost same extra dollop of energy hai. Toh bill proportional hai tum kitne degrees chadhte ho: $Q \propto Δ T$ .

PICTURE. Ek "temperature ladder": 10 rungs chadhne mein kuch heat lagti hai; same ladder ke 20 rungs chadhne mein do guni lagti hai. Heat woh energy ka area hai jo climb ke neeche pour hoti hai.

Step 5 — Constant ka naam: specific heat $c$ , aur law $Q = m c Δ T$

KYA HAI. " $\propto$ " ko " $=$ " mein badlo ek constant insert karke. Woh constant, $c$ , specific heat capacity hai: joules jo is particular material ki 1 kg ko 1 K warm karne ke liye chahiye.

$c$ KYUN, KOI UNIVERSAL NUMBER KYUN NAHI. Water warm hone ka resist karta hai (bada $c$ ); metal aasaani se warm hota hai (chhota $c$ ). Proportionality nahi jaanti tum kaunsa material use kar rahe ho — $c$ woh fingerprint hai jo yeh information carry karta hai. Units seedha nikalta hai: $m Δ T$ (units $kg \cdot K$ ) ko joules banane ke liye, $c$ ko $J kg^{- 1} K^{- 1}$ carry karna hoga.

PICTURE. Same heat bucket equal masses of water vs copper mein daalo: water thermometer muskil se hilta hai, copper ka upar shoot karta hai. "Temperature vs heat added" ka slope chhote- $c$ metal ke liye steeper hai.

Common mistake Hidden assumption: har body ka EK temperature hai (lumped)

Humne quietly kya assume kiya: ki har instant mein ek poora block ek single temperature $T$ par hota hai — koi hot skin aur cold core nahi, andar koi gradient nahi. Tabhi ek number $T$ (aur ek $Δ T$ ) poori object ko $Q = m c Δ T$ mein describe kar sakta hai, aur baad mein ek $T_{1}$ , ek $T_{2}$ cup mein do bodies ko. Ye mostly kyun kaam karta hai: agar hum mixture ko stir karein aur wait karein, ya pieces chhote aur well-conducting hon, toh heat har body ke through spread hoti hai bodies ke beech cross karne se kaafi tez, toh internal differences even out ho jaate hain. Yeh "lumped" (well-mixed) picture hai. Ye kab toot ta hai: ek bade, poorly-conducting body ke liye jo briefly warm kiya gaya ho, andar bahar se lag karta hai — tab tumhe Conduction Convection Radiation ka flow analysis chahiye, single $T$ nahi. Neeche ke har step mein assume kiya gaya hai ki har body ek uniform temperature par lumped hai.

Step 6 — Sealed cup: heat lost = heat gained

KYA HAI. Do objects ko ek insulated cup mein saath rakho. Algebra mein unhe alag rakhne ke liye hum ==subscripts $1$ aur $2$ se label karte hain==: hot body ki mass $m_{1}$ , specific heat $c_{1}$ , starting temperature $T_{1}$ hai; cold body ki $m_{2}, c_{2}, T_{2}$ hai. (Subscript sirf ek name tag hai — " $m_{1}$ " padho "body 1 ki mass".) Hot wale se jo energy nikle uske paas cold wale ke siwa kahin jaane ki jagah nahi hai. Ye tab change karna band karte hain jab dono ek common final temperature $T_{f}$ tak pahunch jaate hain ( $f$ subscript "final" ke liye).

EQUAL KYUN, SIRF "RELATED" KYUN NAHI. Cup insulated hai — room mein koi energy escape nahi hoti. Conservation of Energy ke zariye, hot body jo bhi joule shed karta hai woh joule cold body absorb karti hai: $Q_{lost by hot} = Q_{gained by cold} .$

PICTURE. Ek thermos. Red arrows hot block se nikle; same red arrows cold block mein jaayein. Ek closed loop — andar total energy constant hai.

Step 7 — Meeting point $T_{f}$ ke liye solve karna

KYA HAI. Hamare paas ek equation hai, ek unknown ( $T_{f}$ ). Expand karo aur $T_{f}$ terms ko ek side mein collect karo.

YE ALGEBRA MOVES KYUN. Hum $T_{f}$ akela chahte hain. Brackets multiply out karne se tod ta hai; $T_{f}$ gather karne se hum use factor karke free divide kar sakte hain.

Dono brackets multiply out karo:

$m_{1} c_{1} T_{1} - m_{1} c_{1} T_{f} = m_{2} c_{2} T_{f} - m_{2} c_{2} T_{2}$

Dono $T_{f}$ terms ko right mein, dono constants ko left mein le jaao:

$m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} = m_{2} c_{2} T_{f} + m_{1} c_{1} T_{f}$

Ab key factoring step. Right side par, $T_{f}$ dono terms mein shared factor ke roop mein aata hai — $m_{1} c_{1} T_{f}$ aur $m_{2} c_{2} T_{f}$ dono "(koi number) $\times T_{f}$ " hain. Distributive law ( $a T_{f} + b T_{f} = (a + b) T_{f}$ ) us common $T_{f}$ ko ek single bracket ke bahar kheenchne deta hai:

$m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} = (m_{1} c_{1} + m_{2} c_{2}) T_{f}$

Aakhir mein dono sides ko bracket $(m_{1} c_{1} + m_{2} c_{2})$ se divide karo — allowed hai kyunki yeh ek positive number hai (masses aur specific heats positive hote hain) — $T_{f}$ isolate karne ke liye:

PICTURE. Ek see-saw: $T_{1}$ aur $T_{2}$ sides par baithe hain, aur har object ka $m c$ us end par push karne wala weight hai. $T_{f}$ balance point hai — hamesha dono ke beech, bhaari $m c$ ki taraf khicha hua.

Step 8 — Saare edge cases (taaki kuch surprise na kare)

KYA HAI. Woh corners check karo jahan formula misbehave kar sakta hai.

KYUN. Ek result jis par tum trust karo usse apne extremes survive karne chahiye. Aao har dial ko ek limit tak push karein aur see-saw padho.

Dono sides par equal ( $m_{1} c_{1} = m_{2} c_{2}$ ): weights equal ⇒ balance point plain average hai, $T_{f} = \frac{T _{1} + T _{2}}{2}$ . (Parent ke forecast se match karta hai: 80 °C + 20 °C ⇒ 50 °C.)
Ek side bahut bhaari ( $m_{1} c_{1} ≫ m_{2} c_{2}$ ): see-saw almost hot end ke paas pivot karta hai ⇒ $T_{f} \to T_{1}$ . Ek tiny cold drop ek giant hot ocean ko muskil se cool karta hai.
Same starting temperature ( $T_{1} = T_{2} = T$ ): exchange karne ke liye kuch nahi, $T_{f} = T$ . Plug in karo aur formula exactly $T$ return karta hai.
Vanishing heat weight wali ek body ( $m_{2} c_{2} \to 0$ , e.g. uska mass shrink hokar kuch nahi ho jaata): woh term top aur bottom par vanish ho jaata hai ⇒ $T_{f} \to T_{1}$ . Ek "absent" body temperature shift nahi kar sakti. Caveat: $c_{2} = 0$ exactly set karna sirf ek mathematical degeneracy hai — koi real material ka zero specific heat nahi hota, toh is case ko physical limit "heat weight negligibly small" ke roop mein padho, genuine substance ke roop mein nahi.
Sanity guard: $T_{f}$ kabhi bhi $[T_{2}, T_{1}]$ se bahar nahi ja sakta. Do positive-weighted numbers ka weighted average hamesha unke beech hota hai. Agar tumhara nahi hai, toh tumne "lost" aur "gained" ka sign swap kar diya (parent ka [!mistake] dekho is par).

PICTURE. Chaar mini see-saws dikhate hain: equal weights (centred), lopsided (hot ke paas pivot), same temperature (flat, koi tilt nahi), aur vanishing heat weight (sirf ek weight bacha).

Recall Equal-mass forecast khud check karo

Equal water masses at 80 °C aur 20 °C, $T_{f}$ kya denge? ::: $50$ °C — plain average, kyunki $m_{1} c_{1} = m_{2} c_{2}$ . Bada hot ocean, tiny cold drop — $T_{f}$ kis value ke paas? ::: $T_{1}$ (ocean) ke paas, kyunki uska $m c$ dominate karta hai.

Ek-picture summary

Final figure ek naya visual insight add karta hai jo pehle ke steps ne kabhi nahi draw kiya: yeh cup ke andar do bodies ki actual temperature-vs-time cooling/warming curves overlay karta hai, dikhata hai ki woh ek doosre ki taraf race kar rahe hain aur exactly see-saw balance point $T_{f}$ par milte hain. Poori derivation un do curves ke collision ki kahani hai: jiggle → heat in ( $Q = m c Δ T$ ) → cup ke andar conserve karo → curves $T_{f}$ par milte hain.

Recall Feynman retelling — plain words mein poora walkthrough

Atoms ko balls on springs socho, hamesha jiggling — woh jiggle hi temperature hai. Unhe zyada jiggle karne ke liye tumhe energy daalni padti hai; woh energy-in-transit heat hai, $Q$ . Do gune zyada atoms ko do guni heat chahiye (yeh mass $m$ hai); do guna temperature climb karne ke liye do guni heat chahiye (yeh $Δ T$ hai); aur har material ka apna price tag hai, $c$ (jise hum range par steady treat karte hain, har body poori tarah ek temperature par warm ho ke). Teeno multiply karo aur $Q = m c Δ T$ milta hai. Ab ek hot block aur cold block ko ek thermos mein seal karo jahan se koi energy leak na ho. Hot wala jiggle shed karta hai, cold wala soak karta hai, aur kyunki kuch escape nahi hota, joules lost equals joules gained. Woh do $m c Δ T$ expressions equal set karo, solve karo us temperature ke liye jis par woh finally agree karte hain — shared $T_{f}$ factor out karo aur divide karo — aur ek see-saw nikalta hai: final temperature do starting temperatures ka average hai, har object ke "heat weight" $m c$ se weighted. Bhaara heat-weight jeet ta hai, aur jawaab hamesha kahin do starting temperatures ke beech land karta hai — kabhi bahar nahi. Yahi calorimetry hai, shuru se aakhir tak.

Connections

1.7.04 Specific heat capacity — calorimetry (Hinglish)
Heat and Internal Energy
First Law of Thermodynamics
Conservation of Energy
Latent Heat and Phase Change
Conduction Convection Radiation
Molar Heat Capacity of Gases — Cp and Cv