Specific heat capacity — calorimetry: Worked examples

Before we start, one reminder about the four quantities that do all the work:

The scenario matrix

Every problem this topic can pose is one of these cells. Each row is a "case class"; the last column names the example that clears it.

#	Case class	What is special / the trap	Cleared by
1	Plain heating, one body, $Δ T > 0$	straight $Q = m c Δ T$ , heat in	Ex 1
2	Plain cooling, one body, $Δ T < 0$	sign of $Q$ flips (heat out)	Ex 2
3	Two bodies mix, find $T_{f}$	weighted average, must land between	Ex 3
4	Mix, find unknown $c$	rearrange lost = gained	Ex 4
5	Calorimeter cup included	add cup's $m c$ to the gaining side	Ex 5
6	Degenerate: same start temps	$Δ T = 0 \Rightarrow$ no heat flows	Ex 6
7	Limiting case: one $m c ≫$ other	$T_{f}$ pinned near the big body	Ex 7
8	Real-world word problem	translate words → symbols	Ex 8
9	Exam twist: heater power + time	$Q = P t$ replaces the heat source	Ex 9

We use $c_{water} = 4186 J kg^{- 1} K^{- 1}$ throughout.

Cell 1 — Plain heating, heat in

Cell 2 — Plain cooling, heat out

Cell 3 — Two bodies mix, find $T_{f}$

Worked example Example 3 · Hot and cold water meet

Mix $0.30 kg$ of water at $70^{\circ} C$ with $0.50 kg$ of water at $10^{\circ} C$ in an insulated cup. Find $T_{f}$ .

Forecast: Both are water (same $c$ ), so $c$ cancels — $T_{f}$ is a mass-weighted average. More cold mass ( $0.50$ vs $0.30$ ) means $T_{f}$ should sit below the plain average of $40^{\circ} C$ , closer to $10^{\circ} C$ .

Step 1 — Write the weighted-average formula. $T_{f} = \frac{m _{1} c _{1} T _{1} + m _{2} c _{2} T _{2}}{m _{1} c _{1} + m _{2} c _{2}}$ Why this step? This is the solved form of "heat lost = heat gained". $T_{f}$ is the balance point of two heat capacities.

Step 2 — Cancel $c$ (both water) and plug in. $T_{f} = \frac{0.30 \times 70 + 0.50 \times 10}{0.30 + 0.50} = \frac{21 + 5}{0.80} = \frac{26}{0.80}$ Why this step? Identical $c$ divides out top and bottom, leaving a pure mass weighting.

Step 3 — Divide. $T_{f} = 32.5^{\circ} C$ Why this step? The single fraction from Step 2 is now just arithmetic; carrying it out gives the actual meeting temperature we were asked for.

Verify: $10 < 32.5 < 70$ ✓ ( $T_{f}$ lies between the two starts). Below the plain average $40^{\circ} C$ ✓ (the larger cold mass pulled it down), exactly as forecast.

Figure — Specific heat capacity — calorimetry

Look at the figure: the two starting temperatures are dots on a number line, and $T_{f}$ is the balance point of a see-saw whose arms are weighted by each $m c$ . The heavier $m c$ side pulls the balance point toward itself.

Cell 4 — Mix, find an unknown $c$

Worked example Example 4 · What metal is this?

A $0.25 kg$ metal block at $90^{\circ} C$ is dropped into $0.40 kg$ of water at $18^{\circ} C$ in an ideal (cup absorbs nothing) calorimeter. The mixture settles at $22^{\circ} C$ . Find $c_{metal}$ .

Forecast: The metal fell $68^{\circ} C$ but the water rose only $4^{\circ} C$ — the metal is "eager" to change temperature, so its $c$ should be small compared with water's $4186$ .

Step 1 — Heat gained by water. $Q_{gained} = m_{w} c_{w} Δ T_{w} = 0.40 \times 4186 \times (22 - 18) = 6697.6 J$ Why this step? The water warmed, so this is the energy it took in.

Step 2 — Set heat lost by metal equal to it. $m_{m} c_{m} (T_{hot} - T_{f}) = Q_{gained}$ $0.25 \times c_{m} \times (90 - 22) = 6697.6$ Why this step? Energy conservation in the sealed cup: the metal's loss is exactly the water's gain.

Step 3 — Solve for $c_{m}$ . $c_{m} = \frac{6697.6}{0.25 \times 68} = \frac{6697.6}{17} = 393.98 J kg^{- 1} K^{- 1}$ Why this step? $c_{m}$ is the only unknown, so we divide both sides of Step 2 by the known factor $0.25 \times 68$ to isolate it — that leaves the material fingerprint we wanted.

Verify: $c_{m} \approx 394 ≪ 4186$ ✓ (small, as forecast — close to copper's $\sim 386$ ). Units: $J / (kg \cdot K) = J kg^{- 1} K^{- 1}$ ✓.

Cell 5 — Include the calorimeter cup

Worked example Example 5 · The cup drinks heat too

Repeat Ex 4, but now the copper cup ( $m_{cal} = 0.12 kg$ , $c_{cal} = 386 J kg^{- 1} K^{- 1}$ ) starts at $18^{\circ} C$ with the water and also ends at $22^{\circ} C$ . Find $c_{metal}$ .

Forecast: The cup also absorbs heat, so the "gained" side grows. To supply more heat, the metal's $c_{m}$ must come out slightly larger than Ex 4's $394$ .

Step 1 — Add the cup to the gaining side. $Q_{gained} = (m_{w} c_{w} + m_{cal} c_{cal}) Δ T$ $= (0.40 \times 4186 + 0.12 \times 386) \times (22 - 18)$ Why this step? The cup is in thermal contact and rises the same $4 K$ ; ignoring it undercounts the heat and biases $c_{m}$ low.

Step 2 — Evaluate. $= (1674.4 + 46.32) \times 4 = 1720.72 \times 4 = 6882.88 J$ Why this step? We collapse the two heat sinks into a single combined heat capacity, then multiply by the shared $4 K$ rise to get the total energy the mixture absorbed.

Step 3 — Balance against the metal and solve. $0.25 \times c_{m} \times (90 - 22) = 6882.88$ $c_{m} = \frac{6882.88}{0.25 \times 68} = \frac{6882.88}{17} = 404.87 J kg^{- 1} K^{- 1}$ Why this step? Same energy-conservation balance as Ex 4 — metal's loss equals everything's gain — then divide by the known factor $0.25 \times 68$ to isolate $c_{m}$ .

Verify: $404.87 > 393.98$ ✓ (larger than Ex 4, exactly as forecast — the extra heat sink forces a bigger $c_{m}$ ). The correction is small because the cup's $m c$ is small versus the water's. ✓

Cell 6 — Degenerate case: equal starting temperatures

Worked example Example 6 · Nothing to see here

Mix $0.30 kg$ of water at $25^{\circ} C$ with $0.90 kg$ of oil ( $c_{oil} = 2000 J kg^{- 1} K^{- 1}$ ) also at $25^{\circ} C$ . Find $T_{f}$ .

Forecast: Both already sit at $25^{\circ} C$ . There is no temperature gap to drive any heat flow — expect $T_{f} = 25^{\circ} C$ , whatever the masses.

Step 1 — Plug into the weighted average. $T_{f} = \frac{m _{1} c _{1} ( 25 ) + m _{2} c _{2} ( 25 )}{m _{1} c _{1} + m _{2} c _{2}} = \frac{25 ( m _{1} c _{1} + m _{2} c _{2} )}{m _{1} c _{1} + m _{2} c _{2}}$ Why this step? Whenever both start temperatures are equal, that common value factors out.

Step 2 — Cancel. $T_{f} = 25^{\circ} C$ Why this step? The identical bracket $(m_{1} c_{1} + m_{2} c_{2})$ appears top and bottom and divides to $1$ , leaving only the common temperature — proof that masses and $c$ 's cannot matter here.

Verify: $Δ T = 0$ for both bodies $\Rightarrow Q = 0$ for each ✓. No energy moves — the system was already in thermal equilibrium. The masses and $c$ 's never mattered, which is exactly the meaning of "degenerate". ✓

Cell 7 — Limiting case: one heat capacity dominates

Worked example Example 7 · A drop of hot metal in an ocean of water

Drop $0.010 kg$ of iron ( $c_{iron} = 450 J kg^{- 1} K^{- 1}$ ) at $200^{\circ} C$ into $5.0 kg$ of water at $20^{\circ} C$ . Find $T_{f}$ .

Forecast: Compare the heat capacities. Iron: $m c = 0.010 \times 450 = 4.5$ . Water: $m c = 5.0 \times 4186 = 20930$ . The water's $m c$ is $\sim 4600 \times$ bigger — so $T_{f}$ should be pinned just barely above $20^{\circ} C$ .

Step 1 — Compute both $m c$ values. $m_{Fe} c_{Fe} = 4.5, m_{w} c_{w} = 20930$ Why this step? $T_{f}$ is weighted by $m c$ ; the ratio tells us who "wins".

Step 2 — Weighted average. $T_{f} = \frac{4.5 \times 200 + 20930 \times 20}{4.5 + 20930} = \frac{900 + 418600}{20934.5} = \frac{419500}{20934.5}$ Why this step? Standard mixing formula, now with wildly unequal weights.

Step 3 — Divide. $T_{f} = 20.0387 \dots \approx 20.04^{\circ} C$ Why this step? Carrying out the division turns the fraction into the actual final temperature; the answer barely clears $20$ because the tiny numerator excess ( $900$ vs $418600$ ) is spread over a huge denominator.

Verify: $20 < 20.04 ≪ 200$ ✓, barely off $20^{\circ} C$ ✓ — the huge- $m c$ water acts as a thermal reservoir, absorbing the iron's heat with almost no temperature change. This is the limiting behaviour as one $m c \to \infty$ : $T_{f} \to$ the big body's temperature. ✓

Cell 8 — Real-world word problem

Worked example Example 8 · Cooling tea with a spoon

You have $0.20 kg$ of tea (treat as water) at $85^{\circ} C$ . You stir in a cold aluminium spoon, mass $0.050 kg$ , $c_{Al} = 900 J kg^{- 1} K^{- 1}$ , at $20^{\circ} C$ . Assume no heat leaks. What temperature do they reach?

Forecast: The tea's $m c = 0.20 \times 4186 = 837.2$ dwarfs the spoon's $0.050 \times 900 = 45$ . So $T_{f}$ stays close to $85^{\circ} C$ — the spoon barely cools your tea.

Step 1 — Translate words to symbols. Hot = tea ( $m_{1} = 0.20, c_{1} = 4186, T_{1} = 85$ ); cold = spoon ( $m_{2} = 0.050, c_{2} = 900, T_{2} = 20$ ). Why this step? The whole difficulty of a word problem is naming things; once labelled it is Ex 3 again.

Step 2 — Weighted average. $T_{f} = \frac{0.20 \times 4186 \times 85 + 0.050 \times 900 \times 20}{0.20 \times 4186 + 0.050 \times 900}$ $= \frac{71162 + 900}{837.2 + 45} = \frac{72062}{882.2}$ Why this step? Heat lost by tea = heat gained by spoon; the formula is the packaged answer.

Step 3 — Divide. $T_{f} = 81.68^{\circ} C$ Why this step? The fraction from Step 2 is pure arithmetic now; dividing gives the actual drinking temperature, which is what the word problem asked for.

Verify: $20 < 81.68 < 85$ ✓, and close to $85$ ✓ (the small spoon barely dents the tea) — matching everyday experience and our forecast. ✓

Cell 9 — Exam twist: heater power and time

Worked example Example 9 · An electric heater

A $60 W$ immersion heater runs for $3.0$ minutes in $0.25 kg$ of water, perfectly insulated. What temperature rise results?

Forecast: $60 W$ for $180 s$ delivers $\sim 10 kJ$ . Water's $c$ is large, so expect a modest rise — a few degrees, not tens.

Step 1 — Heat supplied by the heater. $Q = P t = 60 \times (3.0 \times 60) = 60 \times 180 = 10800 J$ Why this step? Power is energy per second, so $Q = P t$ replaces the "heat source" — this is the only new idea; the rest is $Q = m c Δ T$ . We must convert minutes to seconds first, since watts are per second.

Step 2 — Rearrange $Q = m c Δ T$ for $Δ T$ . $Δ T = \frac{Q}{m c} = \frac{10800}{0.25 \times 4186}$ Why this step? We know the heat in and the water; $Δ T$ is the only unknown, so we divide both sides of $Q = m c Δ T$ by $m c$ to isolate it.

Step 3 — Divide. $Δ T = \frac{10800}{1046.5} = 10.32 K$ Why this step? Carrying out the arithmetic converts the isolated expression into the actual number of degrees the water climbs.

Verify: A rise of $\sim 10 K$ ✓ (modest, as forecast). Units: $J / (kg \cdot J kg^{- 1} K^{- 1}) = K$ ✓. If insulation leaked, the real rise would be less — good sanity boundary. ✓

Recall Which cell was which?

Same start temps → no heat flows ::: Ex 6 (degenerate, $Δ T = 0$ ) One $m c$ enormous → reservoir pins $T_{f}$ ::: Ex 7 (limiting case) Heat source is power × time ::: Ex 9 ( $Q = P t$ ) $T_{f}$ must lie between the two starts ::: every mixing example — a built-in sanity check Cup absorbs heat too, so $c_{m}$ comes out larger ::: 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