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Immersion heater: water's specific heat measured electrically

SPH3U Grade 11 Physics · Energy and Society

An electrical calorimetry rig: a small immersion heater sits in 250 g of water in an insulated calorimeter cup, fed by a 12.0 V bench supply with an ammeter in the line reading a steady 4.25 A. The student stirs gently, runs the heater for exactly 5.00 min by a stopwatch, and records the water climbing from 21.3 °C to 35.2 °C. The accepted specific heat of water, for the follow-up, is 4186 J/(kg·K).

Given
  • V = 12 V — Supply voltage
  • I = 4.25 A — Current through the heater
  • t = 5 min — Heating time
  • m = 250 g — Water in the cup
  • T₁ = 21.3 °C — Water before the run
  • T₂ = 35.2 °C — Water after the run
Determine
  1. (a)the electrical power the heater draws
  2. (b)the energy it delivers in the 5.00 min run
  3. (c)the specific heat of water these readings imply
  4. (d)the further heat, and the further time at this power, to bring the cup to a boil
Step 1 of 5(a) · solve for Power

The meters ARE the calorimeter here: P = VI = 51.0 W, joules per second on tap. This is the whole reason the electrical method displaced the mixing method for precision work — electricity can be metered to four figures while a hot block sheds heat on its way to the water.

Rearranged for P
P=VIP = V I
Your values, in your units
P=(12 V) (4.25 A)P = \left( 12\ \text{V} \right) \, \left( 4.25\ \text{A} \right)
Answer
P=51 WP = 51\ \text{W}

Carried onward at full precision, not this rounded figure.

Open the Electrical Power (P = VI) solver →

Step 2 of 5(b) · solve for Energy

E = Pt, with the stopwatch's 5.00 min becoming 300 s before it multiplies anything. Leaving t in minutes is the classic slip in this lab — it hands the water a sixtieth of the energy and a specific heat of about 73 J/(kg·K), absurd enough that the error announces itself.

Rearranged for E
E=PtE = P t
51 Wcarried from step 1
Your values, in your units
E=(51 W) (5 min)E = \left( 51\ \text{W} \right) \, \left( 5\ \text{min} \right)
Converted to base units
E=(51 W) (300 s)E = \left( 51\ \text{W} \right) \, \left( 300\ \text{s} \right)
Answer
E=15.3 kJE = 15.3\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Step 3 of 5(c) · solve for Specific heat capacity

Assume every metered joule landed in the water: c = Q/(mΔT) with the measured 13.9 K rise. The result comes out about 5% HIGH of the accepted 4186 — and the direction is the diagnosis: some joules warmed the cup, the heater's own body and the air, so the water rose less than the electricity paid for, and the blame lands on c.

Rearranged for cₚ
c=Qm ΔTc = \frac{Q}{m \, \Delta T}
15.3 kJcarried from step 2
Your values, in your units
c=(15,300 J)(250 g) (13.9 C∘)c = \frac{\left( 15{,}300\ \text{J} \right)}{\left( 250\ \text{g} \right) \, \left( 13.9\ \text{C}^{\circ} \right)}
Converted to base units
c=(15,300 J)(0.25 kg) (13.9 C∘)c = \frac{\left( 15{,}300\ \text{J} \right)}{\left( 0.25\ \text{kg} \right) \, \left( 13.9\ \text{C}^{\circ} \right)}
Answer
cp=4.4029 kJ/(kg⋅K)c_p = 4.4029\ \text{kJ/(kg}{\cdot}\text{K)}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 4 of 5(d) · solve for Heat energy

Now the accepted c does the predicting: lifting the cup's 250 g from 35.2 °C to the boil at 100.0 °C is a 64.8 K climb, costing 67.8 kJ. Note which c belongs here — the handbook's, because this part is a forecast, not a measurement.

Rearranged for Q
Q=mcΔTQ = m c \Delta T
Your values, in your units
Q=(250 g) (4,186 J/(kg⋅K)) (64.8 C∘)Q = \left( 250\ \text{g} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 64.8\ \text{C}^{\circ} \right)
Converted to base units
Q=(0.25 kg) (4,186 J/(kg⋅K)) (64.8 C∘)Q = \left( 0.25\ \text{kg} \right) \, \left( 4{,}186\ \text{J/(kg}{\cdot}\text{K)} \right) \, \left( 64.8\ \text{C}^{\circ} \right)
Answer
Q=67.813 kJQ = 67.813\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Sensible Heat (Q = mcΔT) solver →

Step 5 of 5 · solve for Time

The same E = Pt read backwards: 67.8 kJ at 51.0 W is over 22 minutes more — four times longer than the run so far bought 13.9 K. A 51 W heater is a teaching instrument, not a kettle; the kitchen version does the same physics thirty times faster.

Rearranged for t
t=EPt = \tfrac{E}{P}
67.813 kJcarried from step 4
51 Wcarried from step 1
Your values, in your units
t=(67,813.2 J)(51 W)t = \tfrac{\left( 67{,}813.2\ \text{J} \right)}{\left( 51\ \text{W} \right)}
Answer
t=22.161 mint = 22.161\ \text{min}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Therefore the heater draws 51.0 W and delivers 15.3 kJ in five minutes, the readings put water's specific heat at 4403 J/(kg·K) — 5.2% above the accepted 4186, the signature of heat leaking to the cup — and boiling the cup from here would take another 67.8 kJ, some 22.2 min at this power.

Why this order

The chain is a supply line: power from the meters, energy from power and the clock, and only then a specific heat from energy and the thermometer. That order matters because each step is a different instrument's testimony — voltmeter and ammeter first, stopwatch second, thermometer last — and the lab's one physical assumption (all metered joules end in the water) is only invoked at step 3, where it can be named and doubted. The two classic wrecks both live in the units: minutes fed to E = Pt starve the energy sixty-fold, and grams fed to c = Q/(mΔT) inflate the answer a thousand-fold. Both produce numbers so far from 4186 that a student who KNOWS the accepted value catches themselves — which is the quiet argument for knowing accepted values.

The 5.2% excess in part (c) is the honest fingerprint of the apparatus, and its direction is fixed: losses always shrink the measured ΔT, so the electrical method always overstates c a little, just as the mixing method understates it. Joule's 1840s paddle-wheel experiments fought exactly this battle — his "mechanical equivalent of heat" is this lab's c in Victorian clothes — and modern calorimeters still bracket the truth between an electrical heating run and a cooling correction. Parts (d) and (e) then flip the worksheet from measuring to engineering: once c is trusted, P·t = mcΔT sizes every kettle, water heater and coffee machine on earth. At 51 W the boil is 22 minutes away; a 1500 W kettle crosses the same 64.8 K in about 45 s. Same equation, different budget.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.