Specific heat capacity — calorimetry: Worked examples

1.7.4 · D3 · Physics › Thermodynamics › Specific heat capacity — calorimetry

Shuru karne se pehle, ek reminder — yeh char quantities hi saara kaam karti hain:

The scenario matrix

Is topic ke har problem ka ek cell hota hai. Har row ek "case class" hai; aakhri column us example ka naam deta hai jo use clear karta hai.

#	Case class	Kya special hai / trap kya hai	Cleared by
1	Plain heating, ek body, $Δ T > 0$	seedha $Q = m c Δ T$ , heat andar	Ex 1
2	Plain cooling, ek body, $Δ T < 0$	$Q$ ka sign flip hota hai (heat bahar)	Ex 2
3	Do bodies mix, $T_{f}$ nikalo	weighted average, beech mein hona chahiye	Ex 3
4	Mix, unknown $c$ nikalo	lost = gained rearrange karo	Ex 4
5	Calorimeter cup bhi shamil	cup ka $m c$ gaining side mein jodo	Ex 5
6	Degenerate: same start temps	$Δ T = 0 \Rightarrow$ koi heat nahi begi	Ex 6
7	Limiting case: ek $m c ≫$ doosra	$T_{f}$ bade body ke paas pin ho jaata hai	Ex 7
8	Real-world word problem	words ko symbols mein translate karo	Ex 8
9	Exam twist: heater power + time	$Q = P t$ heat source ki jagah aata hai	Ex 9

Poore note mein $c_{water} = 4186 J kg^{- 1} K^{- 1}$ use karenge.

Cell 1 — Plain heating, heat in

Cell 2 — Plain cooling, heat out

Cell 3 — Do bodies mix, $T_{f}$ nikalo

Worked example Example 3 · Garam aur thanda paani milte hain

$0.30 kg$ paani $70^{\circ} C$ pe aur $0.50 kg$ paani $10^{\circ} C$ pe ek insulated cup mein milao. $T_{f}$ nikalo.

Forecast: Dono paani hain (same $c$ ), isliye $c$ cancel ho jaata hai — $T_{f}$ ek mass-weighted average hai. Zyada cold mass ( $0.50$ vs $0.30$ ) matlab $T_{f}$ $40^{\circ} C$ ke plain average se neeche hona chahiye, $10^{\circ} C$ ke kareeb.

Step 1 — Weighted-average formula likho. $T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}$ Yeh step kyun? Yeh "heat lost = heat gained" ka solved form hai. $T_{f}$ do heat capacities ka balance point hai.

Step 2 — $c$ cancel karo (dono paani hain) aur plug in karo. $T_{f} = \frac{0.30 \times 70 + 0.50 \times 10}{0.30 + 0.50} = \frac{21 + 5}{0.80} = \frac{26}{0.80}$ Yeh step kyun? Identical $c$ upar aur neeche divide ho jaata hai, sirf mass weighting bachti hai.

Step 3 — Divide karo. $T_{f} = 32.5^{\circ} C$ Yeh step kyun? Step 2 ka ek fraction ab sirf arithmetic hai; ise carry out karne se actual meeting temperature milti hai jo humne poochi thi.

Verify: $10 < 32.5 < 70$ ✓ ( $T_{f}$ dono starts ke beech hai). Plain average $40^{\circ} C$ se neeche ✓ (badi cold mass ne ise neeche kheencha), bilkul forecast jaisa.

Figure — Specific heat capacity — calorimetry

Figure dekho: do starting temperatures number line pe dots hain, aur $T_{f}$ ek see-saw ka balance point hai jiske arms har $m c$ se weighted hain. Bhaari $m c$ wali side balance point ko apni taraf kheenchti hai.

Cell 4 — Mix, unknown $c$ nikalo

Worked example Example 4 · Yeh kaunsa metal hai?

Ek $0.25 kg$ metal block $90^{\circ} C$ pe $0.40 kg$ paani mein $18^{\circ} C$ pe ek ideal (cup kuch nahi absorb karta) calorimeter mein daala jaata hai. Mixture $22^{\circ} C$ pe settle hota hai. $c_{metal}$ nikalo.

Forecast: Metal $68^{\circ} C$ gira lekin paani sirf $4^{\circ} C$ badhaa — metal temperature change ke liye "eager" hai, isliye uska $c$ paani ke $4186$ se chhota hona chahiye.

Step 1 — Paani ne kitni heat gain ki. $Q_{gained} = m_{w} c_{w} Δ T_{w} = 0.40 \times 4186 \times (22 - 18) = 6697.6 J$ Yeh step kyun? Paani garam hua, isliye yeh woh energy hai jo usne li.

Step 2 — Metal ki heat loss ko uske barabar set karo. $m_{m} c_{m} (T_{hot} - T_{f}) = Q_{gained}$ $0.25 \times c_{m} \times (90 - 22) = 6697.6$ Yeh step kyun? Sealed cup mein energy conservation: metal ka loss exactly paani ka gain hai.

Step 3 — $c_{m}$ solve karo. $c_{m} = \frac{6697.6}{0.25 \times 68} = \frac{6697.6}{17} = 393.98 J kg^{- 1} K^{- 1}$ Yeh step kyun? $c_{m}$ akela unknown hai, isliye Step 2 ke dono sides ko known factor $0.25 \times 68$ se divide karte hain taaki use isolate kar sakein — wahi material fingerprint milta hai jo chahiye tha.

Verify: $c_{m} \approx 394 ≪ 4186$ ✓ (chhota, forecast jaisa — copper ke $\sim 386$ ke kareeb). Units: $J / (kg \cdot K) = J kg^{- 1} K^{- 1}$ ✓.

Cell 5 — Calorimeter cup bhi shamil karo

Worked example Example 5 · Cup bhi heat peeta hai

Ex 4 repeat karo, lekin ab copper cup ( $m_{cal} = 0.12 kg$ , $c_{cal} = 386 J kg^{- 1} K^{- 1}$ ) paani ke saath $18^{\circ} C$ se shuru hota hai aur bhi $22^{\circ} C$ pe khatam hota hai. $c_{metal}$ nikalo.

Forecast: Cup bhi heat absorb karta hai, isliye "gained" side badi ho jaayegi. Zyada heat dene ke liye, metal ka $c_{m}$ Ex 4 ke $394$ se thoda bada aana chahiye.

Step 1 — Cup ko gaining side mein jodo. $Q_{gained} = (m_{w} c_{w} + m_{cal} c_{cal}) Δ T$ $= (0.40 \times 4186 + 0.12 \times 386) \times (22 - 18)$ Yeh step kyun? Cup thermal contact mein hai aur same $4 K$ badhta hai; ise ignore karna heat ko undercount karta hai aur $c_{m}$ ko low biased kar deta hai.

Step 2 — Calculate karo. $= (1674.4 + 46.32) \times 4 = 1720.72 \times 4 = 6882.88 J$ Yeh step kyun? Do heat sinks ko ek combined heat capacity mein collapse karte hain, phir shared $4 K$ rise se multiply karte hain taaki mixture ne jo total energy absorb ki woh mile.

Step 3 — Metal ke against balance karo aur solve karo. $0.25 \times c_{m} \times (90 - 22) = 6882.88$ $c_{m} = \frac{6882.88}{0.25 \times 68} = \frac{6882.88}{17} = 404.87 J kg^{- 1} K^{- 1}$ Yeh step kyun? Ex 4 jaisa hi energy-conservation balance — metal ka loss sab ka gain ke barabar — phir $c_{m}$ isolate karne ke liye known factor $0.25 \times 68$ se divide karo.

Verify: $404.87 > 393.98$ ✓ (Ex 4 se bada, bilkul forecast jaisa — extra heat sink bade $c_{m}$ ko force karta hai). Correction chhoti hai kyunki cup ka $m c$ paani ke mukable chhota hai. ✓

Cell 6 — Degenerate case: equal starting temperatures

Worked example Example 6 · Yahan dekhne ko kuch nahi

$0.30 kg$ paani $25^{\circ} C$ pe aur $0.90 kg$ oil ( $c_{oil} = 2000 J kg^{- 1} K^{- 1}$ ) bhi $25^{\circ} C$ pe milao. $T_{f}$ nikalo.

Forecast: Dono pehle se $25^{\circ} C$ pe hain. Koi temperature gap nahi hai jo heat flow drive kare — expect karo $T_{f} = 25^{\circ} C$ , masses kuch bhi hon.

Step 1 — Weighted average mein plug in karo. $T_{f} = \frac{m _{1} c _{1} ( 25 ) + m _{2} c _{2} ( 25 )}{m _{1} c _{1} + m _{2} c _{2}} = \frac{25 ( m _{1} c _{1} + m _{2} c _{2} )}{m _{1} c _{1} + m _{2} c _{2}}$ Yeh step kyun? Jab bhi dono start temperatures equal hoon, woh common value factor out ho jaati hai.

Step 2 — Cancel karo. $T_{f} = 25^{\circ} C$ Yeh step kyun? Identical bracket $(m_{1} c_{1} + m_{2} c_{2})$ upar aur neeche aata hai aur $1$ mein divide ho jaata hai, sirf common temperature bachti hai — proof ki masses aur $c$ yahaan matter nahi kar sakte.

Verify: Dono bodies ke liye $Δ T = 0$ $\Rightarrow$ har ek ke liye $Q = 0$ ✓. Koi energy nahi chalti — system pehle se thermal equilibrium mein tha. Masses aur $c$ kabhi matter hi nahi karte, yahi "degenerate" ka matlab hai. ✓

Cell 7 — Limiting case: ek heat capacity dominate karta hai

Worked example Example 7 · Paani ke samandar mein garm metal ki ek boond

$0.010 kg$ iron ( $c_{iron} = 450 J kg^{- 1} K^{- 1}$ ) $200^{\circ} C$ pe $5.0 kg$ paani $20^{\circ} C$ pe daalo. $T_{f}$ nikalo.

Forecast: Heat capacities compare karo. Iron: $m c = 0.010 \times 450 = 4.5$ . Water: $m c = 5.0 \times 4186 = 20930$ . Paani ka $m c$ $\sim 4600 \times$ bada hai — isliye $T_{f}$ $20^{\circ} C$ se bas thoda sa upar hona chahiye.

Step 1 — Dono $m c$ values compute karo. $m_{Fe} c_{Fe} = 4.5, m_{w} c_{w} = 20930$ Yeh step kyun? $T_{f}$ $m c$ se weighted hota hai; ratio batata hai kaun "jeetta" hai.

Step 2 — Weighted average. $T_{f} = \frac{4.5 \times 200 + 20930 \times 20}{4.5 + 20930} = \frac{900 + 418600}{20934.5} = \frac{419500}{20934.5}$ Yeh step kyun? Standard mixing formula, ab bahut unequal weights ke saath.

Step 3 — Divide karo. $T_{f} = 20.0387 \dots \approx 20.04^{\circ} C$ Yeh step kyun? Division carry out karna fraction ko actual final temperature mein convert karta hai; answer $20$ se bahut kam clearance leta hai kyunki chhota numerator excess ( $900$ vs $418600$ ) ek bade denominator pe spread hota hai.

Verify: $20 < 20.04 ≪ 200$ ✓, barely $20^{\circ} C$ se upar ✓ — bade- $m c$ wala paani ek thermal reservoir ki tarah act karta hai, iron ki heat ko almost bina temperature change ke absorb kar leta hai. Yeh limiting behaviour hai jab ek $m c \to \infty$ : $T_{f} \to$ bade body ka temperature. ✓

Cell 8 — Real-world word problem

Worked example Example 8 · Spoon se chai thandi karo

Tumhare paas $0.20 kg$ chai (paani maano) $85^{\circ} C$ pe hai. Tum ek thanda aluminium spoon, mass $0.050 kg$ , $c_{Al} = 900 J kg^{- 1} K^{- 1}$ , $20^{\circ} C$ pe, hilate ho. Maan lo koi heat leak nahi. Woh kaunse temperature pe aate hain?

Forecast: Chai ka $m c = 0.20 \times 4186 = 837.2$ spoon ke $0.050 \times 900 = 45$ se bahut bada hai. Isliye $T_{f}$ $85^{\circ} C$ ke kareeb rehta hai — spoon tumhari chai ko bahut kam thanda karta hai.

Step 1 — Words ko symbols mein translate karo. Garam = chai ( $m_{1} = 0.20, c_{1} = 4186, T_{1} = 85$ ); thanda = spoon ( $m_{2} = 0.050, c_{2} = 900, T_{2} = 20$ ). Yeh step kyun? Word problem ki puri mushkil cheezein naam dene mein hai; ek baar label ho jaaye toh yeh phir Ex 3 hai.

Step 2 — Weighted average. $T_{f} = \frac{0.20 \times 4186 \times 85 + 0.050 \times 900 \times 20}{0.20 \times 4186 + 0.050 \times 900}$ $= \frac{71162 + 900}{837.2 + 45} = \frac{72062}{882.2}$ Yeh step kyun? Chai ki heat loss = spoon ka heat gain; formula packaged answer hai.

Step 3 — Divide karo. $T_{f} = 81.68^{\circ} C$ Yeh step kyun? Step 2 ka fraction ab pure arithmetic hai; divide karne se actual peene ka temperature milta hai, jo word problem ne poochha tha.

Verify: $20 < 81.68 < 85$ ✓, aur $85$ ke kareeb ✓ (chhota spoon chai mein barely fark dalta hai) — rozmarra ke experience aur forecast se match karta hai. ✓

Cell 9 — Exam twist: heater power aur time

Worked example Example 9 · Ek electric heater

Ek $60 W$ immersion heater $3.0$ minutes tak $0.25 kg$ paani mein, perfectly insulated, chalta hai. Kaunsa temperature rise hota hai?

Forecast: $60 W$ $180 s$ ke liye $\sim 10 kJ$ deliver karta hai. Paani ka $c$ bada hai, isliye modest rise expect karo — kuch degrees, tens nahi.

Step 1 — Heater ne kitni heat supply ki. $Q = P t = 60 \times (3.0 \times 60) = 60 \times 180 = 10800 J$ Yeh step kyun? Power energy per second hai, isliye $Q = P t$ "heat source" ki jagah leta hai — yahi akela naya idea hai; baaki $Q = m c Δ T$ hi hai. Pehle minutes ko seconds mein convert karna zaroori hai, kyunki watts per second hain.

Step 2 — $Q = m c Δ T$ ko $Δ T$ ke liye rearrange karo. $Δ T = \frac{Q}{m c} = \frac{10800}{0.25 \times 4186}$ Yeh step kyun? Heat in aur paani pata hai; $Δ T$ akela unknown hai, isliye $Q = m c Δ T$ ke dono sides ko $m c$ se divide karte hain ise isolate karne ke liye.

Step 3 — Divide karo. $Δ T = \frac{10800}{1046.5} = 10.32 K$ Yeh step kyun? Arithmetic carry out karna isolated expression ko actual number of degrees mein convert karta hai jitne paani chadha.

Verify: $\sim 10 K$ ki rise ✓ (modest, forecast jaisi). Units: $J / (kg \cdot J kg^{- 1} K^{- 1}) = K$ ✓. Agar insulation leak hoti, real rise kam hoti — acha sanity boundary. ✓

Recall Kaun sa cell kaun sa tha?

Same start temps → koi heat nahi begi ::: Ex 6 (degenerate, $Δ T = 0$ ) Ek $m c$ bahut bada → reservoir $T_{f}$ pin karta hai ::: Ex 7 (limiting case) Heat source power × time hai ::: Ex 9 ( $Q = P t$ ) $T_{f}$ dono starts ke beech hona chahiye ::: har mixing example — ek built-in sanity check Cup bhi heat absorb karta hai, isliye $c_{m}$ bada aata hai ::: Ex 5 vs Ex 4

Connections

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