1.7.4
Alag-alag materials heat ko alag tarah "soak up" karte hain. 1 kg paani aur 1 kg copper mein same energy daalo, toh copper bahut zyada garam ho jaata hai. Specific heat capacity is "stubborness" ko measure karta hai: 1 kg substance ka temperature 1 °C badhane ke liye kitni energy chahiye. Calorimetry bas energy ki bookkeeping hai — hot cheezein jo heat release karti hain woh exactly cold cheezein absorb karti hain, kyunki energy gayab nahi ho sakti.
Definition Specific heat capacitySpecific heat capacity c 1 kg substance ka temperature ==1 K (= 1 °C)== badhane ke liye chahiye.
Units: J kg − 1 K − 1 c water ≈ 4186 J kg − 1 K − 1
Definition Heat capacity (poore object ki)C = m c poore object ka temperature 1 K badhane ke liye chahiye heat. Units: J K − 1
Definition Water equivalentPaani ka woh mass w w c water = m c w = c water m c
Intuition Sealed cup mein energy conservationInsulated calorimeter mein koi heat bahar nahi jaati. Toh hot body jo bhi heat release karta hai woh exactly cold body absorb karta hai:
Heat lost by hot = Heat gained by cold
Worked example Compute karne se pehle, predict karo!1 kg paani 80 °C par aur 1 kg paani 20 °C par mix karo. Forecast: equal masses, equal c T f Verify: T f = 4186 + 4186 1 ⋅ 4186 ⋅ 80 + 1 ⋅ 4186 ⋅ 20 = 50
Worked example Example 1 — BasicQ = m c Δ T Q Q = m c Δ T = 0.50 × 4186 × ( 75 − 25 ) Yeh step kyun? Δ T = 50 Q = 0.50 × 4186 × 50 = 1.05 × 1 0 5 J ≈ 104.6 kJ
Worked example Example 2 — Unknown specific heat nikalnaEk 0.20 kg metal block 100 °C par ideal calorimeter mein 0.30 kg paani 20 °C par dala. Final temp 25 °C. c metal Q w = 0.30 × 4186 × ( 25 − 20 ) = 6279 J Yeh step kyun? Paani garam hua, toh usne itna gain kiya.
Metal ne jo heat lose ki = paani ne jo heat gain ki:
0.20 × c m × ( 100 − 25 ) = 6279 Yeh step kyun? Energy conservation: metal ka loss = paani ka gain.
c m = 0.20 × 75 6279 = 418.6 J kg − 1 K − 1
Worked example Example 3 — Calorimeter ko include karna (water equivalent)Ex 2 jaisa hi lekin copper calorimeter ka mass 0.10 kg hai, c cal = 386 paani aur cup dono heat gain karte hain.
Q gained = ( m w c w + m cal c cal ) Δ T = ( 0.30 ⋅ 4186 + 0.10 ⋅ 386 ) ( 5 ) Yeh step kyun? Cup thermal contact mein hai, toh woh bhi heat absorb karta hai — ise ignore karna gain ko underestimate karta hai.
Q gained = ( 1255.8 + 38.6 ) × 5 = 6472 J c m = 0.20 × 75 6472 = 431.5 J kg − 1 K − 1
Common mistake "Temperatures ko °C se pehle kelvin mein convert karna zaroori hai."Kyun sahi lagta hai: thermodynamics formulas mein aksar absolute temperature chahiye.
Fix: Q = m c Δ T difference use karta hai. 50 °C ka gap is 50 K ka gap, kyunki dono scales mein sirf ek constant offset (273.15) ka fark hai. Δ T
Common mistake "Calorimeter cup bhi heat absorb karta hai — yeh bhool jaana."Kyun sahi lagta hai: cup woh substance nahi hai jis par tumhara dhyan hai.
Fix: cup mixture ke contact mein hai aur uska temperature bhi badalta hai. Apna m c Δ T c
T f Kyun sahi lagta hai: tumne mechanically equation solve ki aur number par trust kiya.
Fix: T f weighted average hai, toh yeh T 1 T 2 beech mein hona chahiye. Agar nahi hai, toh signs recheck karo — shayad tumne "lost" aur "gained" ulta kar diya.
Recall Feynman: ek 12-saal ke bachche ko samjhaoSocho do buckets ko same stove par garam kar rahe ho — ek mein paani, ek mein sand. Sand jaldi hot ho jaata hai; paani mein bahut time lagta hai. Paani temperature change karne mein "lazy" hai; woh bahut saari heat store kar sakta hai bina zyada garam hue. Woh laziness hi uski specific heat capacity hai. Ab hot sand ko thande paani mein sealed thermos mein daalo: sand thanda padta hai aur paani garam hota hai jab tak dono beech mein na mil jaayein. Sand jo heat deta hai woh exactly paani jo heat leta hai ke barabar hai — kuch bhi bahar nahi jaata. Woh exchange hi calorimetry hai, aur yeh humein hidden numbers nikalne mein madad karta hai jaise metal ki specific heat.
Mnemonic Formula aur bookkeeping yaad rakho"Q = my CD-Tape" → Q = m c Δ T "HOT loses, COLD gains, TOTAL stays" (energy cup mein conserve hoti hai).
Specific heat capacity c 1 kg substance ka temperature 1 K badhane ke liye chahiye heat; units J kg⁻¹ K⁻¹.
Heat–temperature equation batao. Q = m c Δ T
Q ∝ m ∝ Δ T Do guna mass = do guna atoms ko heat karna; do guna rise = do guna energy add karna.
Q = m c Δ T Yeh temperature ka difference use karta hai; 1 °C ka gap 1 K ke gap ke barabar hai.
Calorimetry ka principle batao. Insulated system mein, hot body ki heat lost = cold body ki heat gained (energy conservation).
Do-body mixture ka final temperature? T f = m 1 c 1 + m 2 c 2 m 1 c 1 T 1 + m 2 c 2 T 2
Heat capacity C c C = m c c
Water equivalent kya hai? Paani ka woh mass jo same Δ T w = m c / c water
Calorimetry mein common omission jo results bigaad deti hai? Yeh bhool jaana ki calorimeter cup bhi heat absorb karta hai; uska m c Δ T
Kya T f Haan — yeh weighted average hai; bahar hona matlab sign error hai.
m1 c1 T1-Tf = m2 c2 Tf-T2
