Kettle on a balance: latent heat of vaporization from mass loss

SPH3U Grade 11 Physics · Energy and Society

A student sets an open-lidded kettle on a kitchen balance, brings it to a full rolling boil, and only then starts the measurement: a plug-in power meter at the wall reads a steady 1440 W, and over exactly 4.00 min of hard boil the balance falls from 1462.0 g to 1312.0 g — 150.0 g gone as steam. The handbook latent heat of vaporization of water, for the audit, is 2257 kJ/kg.

Given
  • P = 1440 WMetered power at the wall
  • t = 4 minTimed interval at full boil
  • Δm = 150 gMass lost to steam (1462.0 − 1312.0 g)
  • L_v = 2257 kJ/kgHandbook latent heat of vaporization
Determine
  1. (a)the energy the kettle delivers during the timed boil
  2. (b)the latent heat of vaporization these readings imply
  3. (c)the mass a perfectly insulated kettle would have boiled off with the same energy
Step 1 of 3(a) · solve for Energy

E = Pt with the METERED 1440 W, not the 1500 W on the rating plate — nameplates are maxima at nominal voltage, and the wall meter is the truth. Starting the clock only after the boil is established is the other half of the design: from then on, no joule goes into raising temperature.

Rearranged for E
E=PtE = P t
Your values, in your units
E=(1,440 W)(4 min)E = \left( 1{,}440\ \text{W} \right) \, \left( 4\ \text{min} \right)
Converted to base units
E=(1,440 W)(240 s)E = \left( 1{,}440\ \text{W} \right) \, \left( 240\ \text{s} \right)
Answer
E=345.6 kJE = 345.6\ \text{kJ}

Carried onward at full precision, not this rounded figure.

Open the Electrical Energy (E = Pt) solver →

Step 2 of 3(b) · solve for Specific latent heat

At a rolling boil the thermometer is parked at 100 °C, so every joule buys phase change: L = Q/m. The result reads about 2% above the handbook 2257 kJ/kg, and high is the only direction it can miss — the kettle's walls shed some heat to the room, so the metered E overstates what the steam actually received.

Rearranged for L
L=QmL = \frac{Q}{m}
345.6 kJcarried from step 1
Your values, in your units
L=(345,600 J)(150 g)L = \frac{\left( 345{,}600\ \text{J} \right)}{\left( 150\ \text{g} \right)}
Converted to base units
L=(345,600 J)(0.15 kg)L = \frac{\left( 345{,}600\ \text{J} \right)}{\left( 0.15\ \text{kg} \right)}
Answer
L=2.304 MJ/kgL = 2.304\ \text{MJ/kg}

Carried onward at full precision, not this rounded figure.

Open the Latent Heat solver →

Step 3 of 3(c) · solve for Mass changing phase

The audit, run the other way: a lossless kettle spending all 345.6 kJ at the handbook rate would have boiled off 153.1 g. The balance saw 150.0 g. The 3.1 g gap is the wall losses and the fine spray that escapes as droplets rather than vapour — about 2% of the water, and the whole error budget of the lab.

Rearranged for m
m=QLm = \frac{Q}{L}
345.6 kJcarried from step 1
Your values, in your units
m=(345,600 J)(2,257 kJ/kg)m = \frac{\left( 345{,}600\ \text{J} \right)}{\left( 2{,}257\ \text{kJ/kg} \right)}
Converted to base units
m=(345,600 J)(2,257,000 J/kg)m = \frac{\left( 345{,}600\ \text{J} \right)}{\left( 2{,}257{,}000\ \text{J/kg} \right)}
Answer
m=153.12 gm = 153.12\ \text{g}

Carried onward at full precision, not this rounded figure.

Open the Latent Heat solver →

Therefore the kettle metered 345.6 kJ into the water, the balance's 150.0 g of lost steam prices vaporization at 2304 kJ/kg — 2.1% over the handbook 2257 — and a perfect kettle would have boiled off 153.1 g, the 3.1 g shortfall being the loss the foam and spray claimed.

Why this order

The elegance of this lab is what it does NOT need: no thermometer, no calorimeter constant, no mixing. Once the boil is rolling, temperature is pinned at the boiling point and the first law collapses to E = mL — so the chain needs only a wall meter, a clock and a balance, in that order. Part (a) must come first because the energy is the only thing being metered; part (b) divides it by the balance's verdict; part (c) reruns the division with the handbook L to expose the difference as grams rather than percent. The two design decisions in the scenario are the actual physics content: start timing AFTER the boil (or sensible heating contaminates E with joules that moved no mass), and read the wall meter, not the rating plate (or the error is built in before the water is).

The number itself deserves awe: 2257 kJ/kg is nearly seven times the 334 kJ/kg that melting cost in the ice-cube chain, and five times the energy needed to heat the same water from 0 to 100 °C. That enormity is why a kettle takes seconds to climb the last degree and minutes to boil dry, why sweating is the body's most powerful cooling instrument, and why a steam burn injures so much worse than boiling water — every condensing gram hands back its 2.26 kJ on contact. It is also why the measured value always errs high in this lab and how to shrink the error: lag the kettle, shield the spray, lengthen the run so the fixed losses dilute. Doubling the timed interval roughly halves the percentage miss — the cheapest precision upgrade in experimental physics.

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