Specific heat capacity — calorimetry: Exercises

Throughout we use the two tools built in the parent note:

Level 1 — Recognition

Recall Solution L1.1

WHAT we want: $Q$ , given $m$ , $c$ , $Δ T$ . Straight use of $Q = m c Δ T$ . WHY: the substance only warms (no melting/boiling), so this single formula covers it. $Δ T = 45 - 15 = 30 K$ (A gap in °C equals the same gap in K.) $Q = 2.0 \times 4186 \times 30 = 251160 J \approx 251 kJ$

Recall Solution L1.2

WHAT: $C$ = heat to warm the whole object by 1 K, so mass is baked in. WHY: $C = m c$ — the "per kg" is removed by multiplying by the kilograms present. $C = m c = 3.0 \times 450 = 1350 J K^{- 1}$

Recall Solution L1.3

WHAT: $w$ such that $w c_{water} = m c_{iron}$ (same heat capacity). WHY: water equivalent replaces the object with an "equally lazy" lump of water. $w = \frac{m c}{c _{water}} = \frac{1350}{4186} = 0.3225 kg$

Level 2 — Application

Recall Solution L2.1

WHAT: heat leaving the water, so we expect $Q < 0$ (heat out). $Δ T = 30 - 90 = - 60 K$ $Q = 0.40 \times 4186 \times (- 60) = - 100464 J$ WHAT IT MEANS: the minus sign says $\approx 100 kJ$ flowed out. The magnitude of heat released is $100464 J \approx 100.5 kJ$ .

Recall Solution L2.2

WHAT: solve $Q = m c Δ T$ for $Δ T$ . WHY divide by $m c$ : $m c$ is the object's heat capacity $C$ ; dividing energy by "energy per degree" gives degrees. $Δ T = \frac{Q}{m c} = \frac{5000}{0.25 \times 900} = \frac{5000}{225} = 22.22 K$

Recall Solution L2.3

PREDICT: equal masses, equal $c$ ⇒ plain average ⇒ $40$ °C. VERIFY with the weighted-average formula: $T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}} = \frac{1 \cdot 4186 \cdot 70 + 1 \cdot 4186 \cdot 10}{4186 + 4186}$ The $4186$ cancels top and bottom: $T_{f} = \frac{70 + 10}{2} = 40 °C ✓$

Level 3 — Analysis

Recall Solution L3.1

Figure — the tug-of-war:

Figure — Specific heat capacity — calorimetry

WHAT: hot metal cools

95 \to 22

, cold water warms

18 \to 22

. WHY set them equal: insulated cup ⇒ metal's heat loss = water's heat gain. Heat gained by water:

Q_{w} = 0.20 \times 4186 \times (22 - 18) = 0.20 \times 4186 \times 4 = 3348.8 J

Heat lost by metal:

Q_{m} = 0.15 \times c_{m} \times (95 - 22) = 0.15 \times c_{m} \times 73

Set

Q_{m} = Q_{w}

0.15 \times 73 \times c_{m} = 3348.8 \Rightarrow c_{m} = \frac{3348.8}{10.95} = 305.8 J kg^{- 1} K^{- 1}

Recall Solution L3.2

WHY closer to 80: the 0.30 kg body has the larger heat capacity $m c$ , so it "wins the tug-of-war" and drags $T_{f}$ toward its own temperature. $T_{f} = \frac{0.30 \cdot 4186 \cdot 80 + 0.10 \cdot 4186 \cdot 20}{0.30 \cdot 4186 + 0.10 \cdot 4186}$ Factor out $4186$ : $T_{f} = \frac{0.30 \cdot 80 + 0.10 \cdot 20}{0.30 + 0.10} = \frac{24 + 2}{0.40} = \frac{26}{0.40} = 65 °C$ Indeed $65$ °C is closer to 80 than to 20. ✓

Recall Solution L3.3

WHY not the plain average: equal masses but unequal $c$ — water's larger $c$ gives it more "pull," so $T_{f}$ leans toward water's 25 °C. Heat capacities: water $m c = 0.50 \times 4186 = 2093$ , oil $m c = 0.50 \times 2000 = 1000$ . $T_{f} = \frac{2093 \cdot 25 + 1000 \cdot 75}{2093 + 1000} = \frac{52325 + 75000}{3093} = \frac{127325}{3093} = 41.17 °C$ Note: below the plain average $50$ °C, exactly as predicted (water pulls it down).

Level 4 — Synthesis

Recall Solution L4.1

WHAT changed: now water AND cup both warm from 18 to 22, so both appear on the heat-gained side. WHY: the cup is in thermal contact — ignoring it undercounts the heat absorbed and biases $c_{m}$ low. $Q_{gained} = (m_{w} c_{w} + m_{cu} c_{cu}) Δ T = (0.20 \cdot 4186 + 0.080 \cdot 386) (22 - 18)$ $= (837.2 + 30.88) \times 4 = 868.08 \times 4 = 3472.32 J$ Set equal to heat lost by metal: $0.15 \times c_{m} \times 73 = 3472.32 \Rightarrow c_{m} = \frac{3472.32}{10.95} = 317.1 J kg^{- 1} K^{- 1}$ Compare to L3.1 ( $305.8$ ): including the cup raises the answer, because more heat was truly absorbed than water alone explained.

Recall Solution L4.2

WHY a general weighted average: with three bodies and no heat leaving, the final temperature is the $m c$ -weighted average of all three starting temperatures. $T_{f} = \frac{\sum m _{i} c _{i} T _{i}}{\sum m _{i} c _{i}}$ Heat capacities:

water A: $0.20 \times 4186 = 837.2$ , at 15
water B: $0.10 \times 4186 = 418.6$ , at 60
Al: $0.050 \times 900 = 45$ , at 90 $Numerator = 837.2 \cdot 15 + 418.6 \cdot 60 + 45 \cdot 90 = 12558 + 25116 + 4050 = 41724$ $Denominator = 837.2 + 418.6 + 45 = 1300.8$ $T_{f} = \frac{41724}{1300.8} = 32.07 °C$ Check: lies between 15 and 90 ✓, and the big cold water (837.2) dominates, pulling $T_{f}$ low toward 15–32 range ✓.

Recall Solution L4.3

WHAT: water equivalent turns the cup's $C$ into an equal mass of water. $w = \frac{C _{cup}}{c _{water}} = \frac{50}{4186} = 0.01194 kg$ WHY useful: now everything is "water," so heat capacities scale purely with mass. Treat the cup as $0.01194$ kg of water at 20 °C mixing with $0.10$ kg water at 90 °C: $T_{f} = \frac{0.10 \cdot 90 + 0.01194 \cdot 20}{0.10 + 0.01194} = \frac{9 + 0.2388}{0.11194} = \frac{9.2388}{0.11194} = 82.53 °C$

Level 5 — Mastery

Recall Solution L5.1

WHAT: both start equal, $T_{1} = T_{2} = 40$ . $T_{f} = \frac{m _{1} c _{1} \cdot 40 + m _{2} c _{2} \cdot 40}{m _{1} c _{1} + m _{2} c _{2}} = 40 \cdot \frac{m _{1} c _{1} + m _{2} c _{2}}{m _{1} c _{1} + m _{2} c _{2}} = 40 °C$ Heat flow: each $Δ T = 0$ , so $Q = m c \cdot 0 = 0$ for both — no heat moves. Equal temperatures = thermal equilibrium already; heat only flows across a temperature difference.

Recall Solution L5.2

WHY intuition first: the lake's heat capacity ( $1000 \times 4186$ ) dwarfs the pebble's ( $0.001 \times 800 = 0.8$ ), so $T_{f}$ barely moves from 10 °C. $T_{f} = \frac{1000 \cdot 4186 \cdot 10 + 0.001 \cdot 800 \cdot 100}{1000 \cdot 4186 + 0.001 \cdot 800} = \frac{41860000 + 80}{4186000 + 0.8}$ $T_{f} = \frac{41860080}{4186000.8} = 10.0000... \approx 10.00 °C$ LIMIT: as one body's $m c \to \infty$ , it acts as a thermal reservoir — its temperature sets $T_{f}$ and stays essentially fixed. This is the idealisation behind a "heat bath."

Recall Solution L5.3

WHY it's impossible: $T_{f}$ is a weighted average of 30 and 50, so it must lie between them. $55$ lies outside $[30, 50]$ — a sure sign the student swapped "lost" and "gained" (a sign error). Correct value (equal $m$ , equal $c$ ⇒ plain average): $T_{f} = \frac{30 + 50}{2} = 40 °C$

Recall Solution L5.4

WHAT: unknown is the cold mass $m$ . Same $c$ cancels. Heat lost by hot water = heat gained by cold water: $0.10 \times c_{w} \times (80 - 35) = m \times c_{w} \times (35 - 20)$ Cancel $c_{w}$ : $0.10 \times 45 = m \times 15 \Rightarrow 4.5 = 15 m \Rightarrow m = 0.30 kg$ Check: the cold body is 3× heavier, so $T_{f} = 35$ sits closer to the cold 20 than to hot 80 ✓.

Recall Self-test summary — reveal only after attempting all

L1: $Q = 251$ kJ; $C = 1350$ J/K; $w = 0.323$ kg. L2: released $100.5$ kJ; $Δ T = 22.2$ K; $T_{f} = 40$ °C. L3: $c_{m} = 305.8$ ; $T_{f} = 65$ °C; $T_{f} = 41.17$ °C. L4: $c_{m} = 317.1$ ; $T_{f} = 32.07$ °C; $w = 0.0119$ kg, $T_{f} = 82.53$ °C. L5: $T_{f} = 40$ , $Q = 0$ ; $T_{f} \approx 10.00$ °C; $T_{f} = 40$ °C; $m = 0.30$ kg.

Connections

Conservation of Energy — the law calorimetry rests on.
Heat and Internal Energy — what "heat" $Q$ physically is.
First Law of Thermodynamics — energy bookkeeping generalised.
Latent Heat and Phase Change — when temperature stops rising.
Conduction Convection Radiation — how the heat actually travels between bodies.
Molar Heat Capacity of Gases — Cp and Cv — the per-mole cousin of $c$ .