Question

The burner on an electric stove has a power output of 2.0 kW. A 610 g stainless steel tea kettle is filled with 20℃ water and placed on the already hot burner.

Part A

If it takes 3.0 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. Express your answer using two significant figures.

Answer 1

Write the expression for the heat required to raise the temperature of water -

q = mCΔT,
where q is the heat added or lost,
m is the mass,
C is the specific heat,
and ΔT is the temperature change.

In this case, heat is being added to both the kettle and the water inside it.
therefore -
qtotal = msCsΔTs + mwCwΔTw

we have given in the problem -  
the mass of the steel (0.610 kg)
the specific heat of iron (0.45 kJ/kgK)
the specific heat of water (4.18 kJ/kgK)
the initial temperature of water and steel (20 ºC)
the final temperature of the water, and therefore of the steel (100 ºC)

again, from the given the power and time, we can find the heat added (a kilowatt is kilojoules per second).
2.0 kW * 3.0 min * 60s/min = 360 kJ

Plugging in:
360 kJ = (0.610 kg)(0.45 kJ/kgK)(100 - 20)K + (mw)(4.18 kJ/kgK)(100 - 20)K

=> 360 kJ = 21.96 kJ + 334.4 mw kJ/kg
=> 338.04 kJ = 334.4 mw kJ/kg
=> mw = 338.04 / 334.4 = 1.011 kg

Now, density of water is 1 g/cm³, so the volume is:
m = 1011 g
v = m/d = 1011 g / (1g/cm³) = 1011 cm³

In two significant figures your answer is -

v = 1010 cm³