Question

The burner on an electric stove has a power output of 2.0 kW. A 760 g stainless steel tea kettle is filled with 20∘C water and placed on the already hot burner.

If it takes 2.8 min for the water to reach a boil, what volume of water, in cm3, was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Answer 1

q = mCΔT,

where q is the heat added or lost,

m is the mass,

C is the specific heat,

ΔT is the temperature change.

In this case, heat is being added to both the kettle and the water inside it.

qtotal = msCsΔTs + mwCwΔTw

mass of the steel (0.760 kg)

specific heat of iron (0.45 kJ/kgK)

specific heat of water (4.18 kJ/kgK)

initial temperature of water and steel (20 ºC)

final temperature of the water, and therefore of the steel (100 ºC)

Given the power and time, we can find the heat added (a kilowatt is kilojoules per second).

2 kW * 2.8 min * 60s/min = 336 kJ

336 kJ = (0.760 kg)(0.45 kJ/kgK)(100 - 20)K + (mw)(4.18 kJ/kgK)(100 - 20)K

336 kJ = 27.36 kJ + 334.4 mw kJ/kg

308.64 kJ = 334.4 mw kJ/kg

mw = 0.923 kg

The density of water is 1 g/cm^3, so the volume is:

m = 923 g

v = m/d

v = 923 g / (1g/cm^3)

v = 923 cm^3