The electrical calorimetry rig takes up half the bench: a 12.0 V bench supply, an ammeter wired into the line, a small immersion heater, an insulated calorimeter cup, a thermometer, and a stopwatch. The student measures 250 g of water into the cup and seats the heater so its coil sits fully under the surface — an element run in air burns out in seconds. At switch-on the ammeter settles to a steady 4.25 A and the stopwatch starts in the same motion. The student stirs gently for the whole run, because without stirring the hot water pools around the element and the thermometer reads the pool, not the cup. The heater runs for exactly 5.00 min by the stopwatch before the power is cut, and the record shows the water climbing from 21.3 °C at the start to 35.2 °C at switch-off. The accepted specific heat of water, for the follow-up, is 4,186 J/(kg·K).
Every number in this problem is editable, the material included — change any value below and the whole chain recalculates.
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
(a)the electrical power the heater draws
(b)the energy it delivers in the 5.00 min run
(c)the specific heat of water these readings imply
(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=VI
Your values, in your units
P=(12V)(4.25A)
Answer
P=51W
Carried onward at full precision, not this rounded figure.
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=Pt
51 Wcarried from step 1
Your values, in your units
E=(51W)(5min)
Converted to base units
E=(51W)(300s)
Answer
E=15.3kJ
Carried onward at full precision, not this rounded figure.
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 4,186 — 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 delivered, and the blame lands on c.
Rearranged for cₚ
c=mΔTQ
15.3 kJcarried from step 2
Your values, in your units
c=(250g)(13.9C∘)(15,300J)
Converted to base units
c=(0.25kg)(13.9C∘)(15,300J)
Answer
cp=4.4029kJ/(kg⋅K)
Carried onward at full precision, not this rounded figure.
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=mcΔT
Your values, in your units
Q=(250g)(4,186J/(kg⋅K))(64.8C∘)
Converted to base units
Q=(0.25kg)(4,186J/(kg⋅K))(64.8C∘)
Answer
Q=67.813kJ
Carried onward at full precision, not this rounded figure.
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 produced 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=PE
67.813 kJcarried from step 4
51 Wcarried from step 1
Your values, in your units
t=(51W)(67,813.2J)
Answer
t=22.161min
Carried onward at full precision, not this rounded figure.
Therefore the heater draws 51.0 W and delivers 15.3 kJ in five minutes, the readings put water's specific heat at 4,403 J/(kg·K) — 5.2% above the accepted 4,186, 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 4,186 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 1,500 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.
