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∘C 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.

Answer 1

Using energy conservation

Total Energy required = Total energy supplied

Energy Supplied = Power*time

time = 3 min = 3*60 sec

Es = 2*10^3*3*60 = 3.6*10^5 J

Energy required = Q1 + Q2

Q = m*C*dT

Er = mw*Cw*dT + ms*Cs*dT

dT = 100 - 20 = 80

mw = ?

ms = 610 gm = 0.61 kg

Cw = 4186 J/kg-C

Cs = 450 J/kg-C

Er = mw*4186*80 + 0.61*450*80

Es = Er

3.6*10^5 = mw*4186*80 + 0.61*450*80

mw = (3.6*10^5 - 0.61*450*80)/(4186*80)

mw = 1.009 kg = 1009 g of water

we know that

density of water = 1 gm/cm^3

Volume = mass/density

Volume = 1009 gm/(1 gm/cm^3) = 1009 cm^3