Intuition The big pictureDifferent materials "soak up" heat differently. Pour the same energy into 1 kg of water and 1 kg of copper, and the copper gets much hotter. Specific heat capacity measures this stubbornness: how much energy it takes to warm 1 kg of a substance by 1 °C. Calorimetry is just energy bookkeeping — heat lost by hot things equals heat gained by cold things, because energy can't vanish.
Definition Specific heat capacityThe specific heat capacity c c c 1 kg of a substance by ==1 K (= 1 °C)==.
Units: J kg − 1 K − 1 \text{J kg}^{-1}\,\text{K}^{-1} J kg − 1 K − 1 c water ≈ 4186 J kg − 1 K − 1 c_{\text{water}} \approx 4186\ \text{J kg}^{-1}\text{K}^{-1} c water ≈ 4186 J kg − 1 K − 1
Definition Heat capacity (of a whole object)C = m c C = mc C = m c whole object by 1 K. Units: J K − 1 \text{J K}^{-1} J K − 1
Definition Water equivalentThe mass of water w w w same heat as a given object for the same temperature rise: w c water = m c w c_{\text{water}} = m c w c water = m c w = m c c water w = \dfrac{mc}{c_{\text{water}}} w = c water m c
Intuition Why proportional to mass AND toΔ T \Delta T Δ T
Twice the mass = twice the atoms to shake ⇒ heat ∝ m \propto m ∝ m Twice the temperature rise = twice as much shaking added ⇒ heat ∝ Δ T \propto \Delta T ∝ Δ T The proportionality constant per kilogram is the material's fingerprint, c c c
Intuition Conservation of energy in a sealed cupIn an insulated calorimeter no heat escapes. So whatever heat the hot body releases is exactly the heat the cold body absorbs:
Heat lost by hot = Heat gained by cold \text{Heat lost by hot} = \text{Heat gained by cold} Heat lost by hot = Heat gained by cold
Worked example Before computing, predict!Mix 1 kg water at 80 °C with 1 kg water at 20 °C. Forecast: equal masses, equal c c c T f T_f T f Verify: T f = 1 ⋅ 4186 ⋅ 80 + 1 ⋅ 4186 ⋅ 20 4186 + 4186 = 50 T_f = \frac{1\cdot4186\cdot80 + 1\cdot4186\cdot20}{4186+4186} = 50 T f = 4186 + 4186 1 ⋅ 4186 ⋅ 80 + 1 ⋅ 4186 ⋅ 20 = 50
Worked example Example 1 — BasicQ = m c Δ T Q=mc\Delta T Q = m c Δ T Q Q Q Q = m c Δ T = 0.50 × 4186 × ( 75 − 25 ) Q = mc\Delta T = 0.50 \times 4186 \times (75-25) Q = m c Δ T = 0.50 × 4186 × ( 75 − 25 ) Why this step? Δ T = 50 \Delta T = 50 Δ T = 50 Q = 0.50 × 4186 × 50 = 1.05 × 10 5 J ≈ 104.6 kJ Q = 0.50 \times 4186 \times 50 = 1.05\times10^{5}\ \text{J} \approx 104.6\ \text{kJ} Q = 0.50 × 4186 × 50 = 1.05 × 1 0 5 J ≈ 104.6 kJ
Worked example Example 2 — Find an unknown specific heatA 0.20 kg metal block at 100 °C is dropped into 0.30 kg water at 20 °C in an ideal calorimeter. Final temp 25 °C. Find c metal c_{\text{metal}} c metal Q w = 0.30 × 4186 × ( 25 − 20 ) = 6279 J Q_w = 0.30 \times 4186 \times (25-20) = 6279\ \text{J} Q w = 0.30 × 4186 × ( 25 − 20 ) = 6279 J Why this step? Water warmed up, so it gained this much.
Heat lost by metal = heat gained by water:
0.20 × c m × ( 100 − 25 ) = 6279 0.20 \times c_m \times (100-25) = 6279 0.20 × c m × ( 100 − 25 ) = 6279 Why this step? Energy conservation: metal's loss = water's gain.
c m = 6279 0.20 × 75 = 418.6 J kg − 1 K − 1 c_m = \dfrac{6279}{0.20 \times 75} = 418.6\ \text{J kg}^{-1}\text{K}^{-1} c m = 0.20 × 75 6279 = 418.6 J kg − 1 K − 1
Worked example Example 3 — Including the calorimeter (water equivalent)Same as Ex 2 but the copper calorimeter has mass 0.10 kg, c cal = 386 c_{\text{cal}}=386 c cal = 386 both water and cup gain heat.
Q gained = ( m w c w + m cal c cal ) Δ T = ( 0.30 ⋅ 4186 + 0.10 ⋅ 386 ) ( 5 ) Q_{\text{gained}} = (m_w c_w + m_{\text{cal}}c_{\text{cal}})\,\Delta T = (0.30\cdot4186 + 0.10\cdot386)(5) Q gained = ( m w c w + m cal c cal ) Δ T = ( 0.30 ⋅ 4186 + 0.10 ⋅ 386 ) ( 5 ) Why this step? The cup is in thermal contact, so it also absorbs heat — ignoring it underestimates the gain.
Q gained = ( 1255.8 + 38.6 ) × 5 = 6472 J Q_{\text{gained}} = (1255.8 + 38.6)\times 5 = 6472\ \text{J} Q gained = ( 1255.8 + 38.6 ) × 5 = 6472 J c m = 6472 0.20 × 75 = 431.5 J kg − 1 K − 1 c_m = \dfrac{6472}{0.20\times75} = 431.5\ \text{J kg}^{-1}\text{K}^{-1} c m = 0.20 × 75 6472 = 431.5 J kg − 1 K − 1
Common mistake "Temperatures in °C must be converted to kelvin first."Why it feels right: thermodynamics formulas often need absolute temperature.
Fix: Q = m c Δ T Q=mc\Delta T Q = m c Δ T difference . A gap of 50 °C is a gap of 50 K, because the scales differ only by a constant offset (273.15). Use °C or K freely for Δ T \Delta T Δ T
Common mistake "Forget the calorimeter cup absorbs heat."Why it feels right: the cup isn't the substance you care about.
Fix: the cup is in contact with the mixture and changes temperature too. Add its m c Δ T m c\,\Delta T m c Δ T c c c
T f T_f T f Why it feels right: you mechanically solved an equation and trusted the number.
Fix: T f T_f T f weighted average , so it must lie between T 1 T_1 T 1 T 2 T_2 T 2
Recall Feynman: explain to a 12-year-oldImagine warming up two buckets — one of water, one of sand — on the same stove. The sand gets hot fast; the water takes forever. Water is "lazy" about changing temperature; it can store lots of heat without warming much. That laziness is its specific heat capacity . Now mix hot sand into cool water in a sealed thermos: the sand cools down and the water warms up until they meet in the middle. The heat the sand gives away is exactly the heat the water takes in — nothing leaks out. That swap is calorimetry, and it lets us figure out hidden numbers like a metal's specific heat.
Mnemonic Remember the formula & the bookkeeping"Q = my CD-Tape" → Q = m c Δ T Q = m\,c\,\Delta T Q = m c Δ T "HOT loses, COLD gains, TOTAL stays" (energy is conserved in the cup).
What does specific heat capacity c c c Heat needed to raise 1 kg of a substance by 1 K; units J kg⁻¹ K⁻¹.
State the heat–temperature equation. Q = m c Δ T Q = mc\Delta T Q = m c Δ T .
Why is Q ∝ m Q \propto m Q ∝ m ∝ Δ T \propto \Delta T ∝ Δ T Twice the mass = twice the atoms to heat; twice the rise = twice the energy added.
Why can you use °C instead of K in Q = m c Δ T Q=mc\Delta T Q = m c Δ T It uses a temperature difference ; a 1 °C gap equals a 1 K gap.
State the principle of calorimetry. In an insulated system, heat lost by hot body = heat gained by cold body (energy conservation).
Final temperature of a two-body mixture? T f = m 1 c 1 T 1 + m 2 c 2 T 2 m 1 c 1 + m 2 c 2 T_f = \dfrac{m_1c_1T_1 + m_2c_2T_2}{m_1c_1+m_2c_2} T f = m 1 c 1 + m 2 c 2 m 1 c 1 T 1 + m 2 c 2 T 2 — a heat-capacity-weighted average.
What is heat capacity C C C c c c C = m c C = mc C = m c , for the whole object (J K⁻¹);
c c c is per kg.
What is water equivalent? Mass of water needing the same heat for the same
Δ T \Delta T Δ T :
w = m c / c water w = mc/c_\text{water} w = m c / c water .
Common calorimetry omission that skews results? Forgetting the calorimeter cup also absorbs heat; add its
m c Δ T mc\Delta T m c Δ T .
Must T f T_f T f Yes — it's a weighted average; outside means a sign error.
m1 c1 T1-Tf = m2 c2 Tf-T2
Intuition Hinglish mein samjhoDekho, har material ka apna "swabhav" hota hai garmi ke saath. Specific heat capacity c c c c c c c c c Q = m c Δ T Q = mc\Delta T Q = m c Δ T
Calorimetry sirf ek hisaab-kitaab hai energy ka. Jab garam cheez ko thande paani mein daalo (insulated cup mein), toh garam cheez jitni heat chhodti hai, thandi cheez utni hi heat le leti hai — kuch bahar nahi jaata. Isi se hum likhte hain: heat lost = heat gained , yaani m 1 c 1 ( T 1 − T f ) = m 2 c 2 ( T f − T 2 ) m_1c_1(T_1-T_f) = m_2c_2(T_f-T_2) m 1 c 1 ( T 1 − T f ) = m 2 c 2 ( T f − T 2 )
Do important baatein yaad rakho. Pehli — Δ T \Delta T Δ T difference hai (50 °C ka gap = 50 K ka gap). Kelvin mein convert karne ki tension mat lo. Doosri — calorimeter ka cup bhi heat absorb karta hai, usko bhool gaye toh answer galat aayega; uska m c Δ T mc\Delta T m c Δ T
Aur ek check: T f T_f T f beech mein hoga, kyunki yeh ek weighted average hai (weight = m c mc m c T f T_f T f
