Specific heat capacity — calorimetry: Question bank

1.7.4 · D5 · Physics › Thermodynamics › Specific heat capacity — calorimetry

Reveal format ek line ka hai: statement ::: reasoning. Sirf tab uncover karo jab tum zor se answer de chuko.

Notation key — shuru karne se pehle padho

Neeche har trap same symbols reuse karta hai. Inhe ek baar pin down kar lo taaki koi letter unearned na lage:

Definition Symbols aur sign conventions

$Q$ — heat, energy (joules, J) jo temperature difference ki wajah se kisi body mein andar ya bahar flow karti hai.
$m$ — body ki mass (kg).
$c$ — specific heat capacity (J kg⁻¹ K⁻¹): 1 kg ko 1 K warm karne ke liye kitni heat chahiye. Ek material fingerprint.
$C = m c$ — poore object ki heat capacity (J K⁻¹): mass bake in hai, toh "per kg" nahi.
$Δ T = T_{final} - T_{initial}$ — temperature change, hamesha final minus initial.
Sign rule: us definition ke saath, $Q = m c Δ T$ mein $Q > 0$ ka matlab hai heat andar aayi (body warm hoti hai, $T_{final} > T_{initial}$ ) aur $Q < 0$ ka matlab hai heat bahar gayi (body cool hoti hai). $Δ T$ flip karo aur tum silently har sign flip kar dete ho.
Subscripts bodies ko label karte hain: body 1 ke liye $m_{1}, c_{1}, T_{1}$ ; body 2 ke liye $m_{2}, c_{2}, T_{2}$ ; cup hai $m_{cal}, c_{cal}$ . Shared end temperature hai $T_{f}$ (kabhi "Tf" nahi likhte).

$T_{f}$ ek weighted average kyon hai — woh derivation jis par tum lean karoge

Kaafi traps " $T_{f}$ is a weighted average" quote karte hain. Ye raha kyon, step by step, taaki ye sirf ek slogan na rahe.

Hum kahan se shuru karte hain. Do bodies ek insulated cup mein heat swap karti hain jab tak $T_{f}$ share na kar lein. Body 1 (hot) lose karti hai $Q_{lost} = m_{1} c_{1} (T_{1} - T_{f})$ ; body 2 (cold) gain karti hai $Q_{gained} = m_{2} c_{2} (T_{f} - T_{2})$ . Dono positive quantities ke roop mein likhe hain kyunki $T_{1} > T_{f} > T_{2}$ .

Hum inhe equate kyon karte hain. Koi heat bahar nahi jaati, isliye Conservation of Energy force karta hai lost = gained: $m_{1} c_{1} (T_{1} - T_{f}) = m_{2} c_{2} (T_{f} - T_{2})$

Hum aage kya karte hain — $T_{f}$ solve karo. Expand karo aur $T_{f}$ terms collect karo: $m_{1} c_{1} T_{1} + m_{2} c_{2} T_{2} = (m_{1} c_{1} + m_{2} c_{2}) T_{f}$ $T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}$

YE KAISA DIKHTA HAI. Ye exactly ek weighted average $\frac{w _{1} T _{1} + w _{2} T _{2}}{w _{1} + w _{2}}$ ki shape hai jisme weights $w_{i} = m_{i} c_{i}$ hain — heat capacities. Figure dekho: number line par do temperatures, aur $T_{f}$ ek balance point hai jo zyada bade $m c$ wali body ki taraf khichta hai. Ek balance point hamesha apne do loads ke beech baith ta hai, isliye $T_{f}$ kabhi interval $[T_{2}, T_{1}]$ se bahar nahi ja sakta.

Figure — Specific heat capacity — calorimetry

True ya false — justify karo

Q = m c Δ T

mein °C ki jagah K use karna galat answer deta hai.

False. Formula ek difference

Δ T

use karta hai, aur 50 °C ka gap 50 K ke gap jaisa hi hai (scales mein sirf constant offset 273.15 ka fark hai). Sirf absolute-temperature formulas mein conversion chahiye.

Jis material ka

c

bada ho woh stove par jaldi heat hoti hai.

False. Bada

c

matlab woh temperature change resist karti hai — woh har degree ke liye bahut zyada energy soak karti hai, isliye woh slowly warm hoti hai. Water (bada

c

) sand (chhota

c

) se peeche rehta hai.

Heat capacity

C

aur specific heat

c

alag units mein same quantity hain.

False.

c

per kilogram hai (ek material fingerprint);

C = m c

object ki mass bake in karta hai, toh low-

c

metal ka bada block high-

c

water ke chhote blob se zyada bada

C

rakh sakta hai.

Ideal calorimeter mein mixture ki total internal energy constant rehti hai.

True. "Insulated" ka matlab hai koi heat bahar nahi jaati, isliye hot body ka loss exactly cold body ke gain ke barabar hota hai — total energy conserved hai, Conservation of Energy ke zariye.

Agar mixture mein dono masses double kar do, toh final temperature

T_{f}

bhi double ho jaati hai.

False.

T_{f}

starting temperatures ka ek weighted average hai; dono masses ko 2 se multiply karna fraction ke top aur bottom dono ko 2 se multiply karta hai aur cancel ho jaata hai —

T_{f}

unchanged rehta hai.

Same temperature par do alag materials mein har kilogram mein utni hi heat hoti hai.

False. Temperature aur stored thermal energy alag cheezein hain; jiska

c

zyada bada hai woh same temperature par zyada energy hold karta hai.

Kisi object ka water equivalent joules mein hota hai.

False. Ye ek mass hai (kg) — pani ki woh mass jisko same

Δ T

ke liye same heat chahiye:

w = m c / c_{water}

T_{f}

dono starting temperatures se zyada ho sakta hai agar hot body bahut heavy ho.

False. Weighted average hamesha do inputs ke beech hi land karta hai. Bahar ki value matlab "lost" vs "gained" mein sign error hai.

Error dhundo

"0.20 kg metal 100 °C par hai, pani mein daala, final 25 °C; pani ne

Q_{w} = 6279

J heat gain ki. Toh

c_{m} = \frac{6279}{0.20 \times 25}

." Galti dhundo.

Yahan

6279

J woh heat hai jo pani ne gain ki, jo metal ke lost heat ke barabar hai. Lekin metal ka apna

Δ T

hai

100 - 25 = 75

K, na ki 25. Denominator mein metal ka drop use karna chahiye:

c_{m} = 6279/ (0.20 \times 75)

"Metal se heat lost

= m_{1} c_{1} (T_{f} - T_{1})

." Kya galat hai?

Sign flipped hai. Hot body cool hoti hai, isliye

T_{1} > T_{f}

aur uska loss hai

m_{1} c_{1} (T_{1} - T_{f})

— ek positive quantity. Jaise likha hai woh negative hoga.

"Copper cup ko ignore karte hain kyunki woh measured substance nahi hai." Ye galat kyon hai?

Cup thermal contact mein hai aur pani ke saath warm hota hai,

m_{cal} c_{cal} Δ T

absorb karta hai. Ise omit karna gained heat ko underestimate karta hai aur computed

c

ko skew karta hai.

Δ T = T_{initial} - T_{final}

Q = m c Δ T

." Trap dhundo.

Convention hai

Δ T = T_{final} - T_{initial}

, taaki

Q > 0

ka matlab heat andar aayi. Ise reverse karna silently har result ka sign flip kar deta hai.

"Kyunki dono bodies

T_{f}

reach karti hain, hum

m_{1} c_{1} T_{1} = m_{2} c_{2} T_{2}

set karte hain." Ye nonsense kyon hai?

Tumhe heat exchanged equate karni chahiye, sirf products

m c T

nahi. Sahi balance hai

m_{1} c_{1} (T_{1} - T_{f}) = m_{2} c_{2} (T_{f} - T_{2})

"Pani aur metal mix kiya, dono ke liye water ka

c

use kiya kyunki woh sabse bada hai." Fix karo.

Har substance apna khud ka

c

rakhta hai; heat gained mein

c_{water}

use hota hai, heat lost mein

c_{metal}

use hota hai. Sirf common final temperature

T_{f}

share hoti hai.

Why questions

Q

dono mass aur temperature change ke proportional kyon hai?

Double mass matlab double atoms ko shake karna, aur double temperature rise matlab har atom mein double shaking add karna — dono effects multiply karte hain, jo force karta hai

Q \propto m Δ T

Bade

m c

wala object

T_{f}

ke tug-of-war mein kyon "jeetta" hai?

T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}

mein har body ka weight uski heat capacity

m c

hai; bada

m c

temperature change ko strongly resist karta hai, isliye equilibrium us body ki starting temperature ke paas settle hoti hai.

Real (non-ideal) experiment mein metal ka

c

thoda off kyon aata hai?

Real cups surroundings mein heat leak karte hain aur cup khud kuch absorb karta hai; jab tak dono account nahi hote, measured heat gained galat hoti hai,

c

ko bias karti hai.

Calorimeter ko alag track karne ki jagah "water equivalent" se kyon treat kar sakte hain?

Kyunki heat sirf product

m c Δ T

par depend karti hai; cup ko equivalent mass of water

w = m_{cal} c_{cal} / c_{water}

se replace karna identical heat absorb karta hai aur bookkeeping simplify hoti hai.

80 °C par 1 kg pani ko 20 °C par 1 kg pani mein mix karne par exactly 50 °C kyon milta hai?

Equal masses aur equal

c

weights ko identical banate hain, isliye heat-capacity-weighted average plain arithmetic mean mein collapse ho jaata hai,

(80 + 20) /2

Calorimetry ultimately Conservation of Energy disguise mein kyon hai?

Ek sealed insulated cup bahar energy exchange nahi karta, isliye hot body se nikli koi bhi thermal energy cold body mein reappear karni chahiye — heat lost equals heat gained.

Edge cases

Agar do identical bodies ko same temperature par mix karo toh

T_{f}

kya hai?

Woh usi temperature par rehti hai; dono ke liye

Δ T = 0

, koi heat flow nahi hoti, aur do equal values ka weighted average wahi value hoti hai.

Agar ek body effectively infinite mass ki ho (ek bada reservoir),

T_{f}

ka kya hoga?

Reservoir ka

m c

weights mein completely dominate karta hai, isliye

T_{f}

reservoir ki temperature par pin ho jaata hai — chhoti body simply usi par relax kar leti hai.

Agar mixing ke dauran "final" temperature 0 °C tak pahunch jaaye — kya

Q = m c Δ T

kaafi hai?

Zaruri nahi. Agar phase change (freezing/melting) hoti hai, toh

Q = m c Δ T

sirf state change ke bina change cover karta hai; tumhe balance mein ek latent heat term add karna hoga (next item).

Latent heat define karo aur dikhao ye heat balance mein kahan aata hai.

Latent heat

L

(J kg⁻¹) woh energy hai jo constant temperature par 1 kg ka phase change karne ke liye chahiye; ye extra term

Q = m L

ke roop mein gained/lost side mein enter karta hai, isliye jo body cool bhi hoti hai aur freeze bhi hoti hai woh contribute karti hai

m c Δ T + m L

Kya

T_{f}

exactly kisi ek starting temperature ke barabar ho sakta hai?

Sirf us limit mein jab doosri body ki heat capacity zero ho (

m \to 0

c \to 0

), toh woh kuch absorb/release nahi karti aur equilibrium shift nahi kar sakti.

Agar do bodies pehle se thermal equilibrium mein hain, kya unke beech heat flow hoti hai?

Koi net heat flow nahi hoti jab temperatures equal ho jaati hain — "temperature difference" ka driving force zero hai, isliye system rest par hai, chahe microscopic energy exchange dono taraf continue ho.

Kya

Q = m c Δ T

bade temperature range par exact rehta hai?

Nahi — ye assume karta hai ki

c

constant hai. Bade ranges par

c

temperature ke saath drift karta hai, isliye exact heat hai

Q = \int m c (T) d T

; simple product sirf small-range approximation hai.

Recall One-line self-test

Woh ek reason jisse $T_{f}$ kabhi do starting temperatures ke bahar nahi ja sakta? ::: Ye ek weighted average hai, aur average hamesha apne inputs ke beech hota hai — bahar ki value lost/gained sign error indicate karti hai.

Question bank — Specific heat capacity — calorimetry

Notation key — shuru karne se pehle padho

$T_{f}$ ek weighted average kyon hai — woh derivation jis par tum lean karoge

True ya false — justify karo

Error dhundo

Why questions

Edge cases

Connections

Notation key — shuru karne se pehle padho

Tf​ ek weighted average kyon hai — woh derivation jis par tum lean karoge

True ya false — justify karo

Error dhundo

Why questions

Edge cases

Connections

$T_{f}$ ek weighted average kyon hai — woh derivation jis par tum lean karoge