Question

The burner on an electric stove has a power output of 2.0 kW. A 800 g stainless steel tea kettle is filled with 20∘C water and placed on the already hot burner.

Part A

If it takes 3.4 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

Using energy conservation

Total Energy required = Total energy supplied

Energy Supplied = Power*time

time = 3.4 min = 3.4*60 sec

Power = 2.0 kW

Es = 2*10^3*3.4*60 = 408000 J

Energy required = Q1 + Q2

Q = m*C*dT

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

dT = change in temperature = 100 - 20 = 80

mw = mass of water = ?

ms = mass of aluminium = 800 gm = 0.800 kg

Cw = 4186 J/kg-C

Cs = specific heat of iron = 450 J/kg-C

Er = mw*4186*80 + 0.8*450*80

Es = Er

408000 = mw*4186*80 + 0.8*450*80

mw = (408000 - 0.8*450*80)/(4186*80)

mw = 1.132 kg = 1132 g of water

we know that

density of water = 1 gm/cm^3

Volume = mass/density

Volume = 1132 gm/(1 gm/cm^3) = 1132 cm^3

In two significant figure:

Volume = 1100 cm^3

"Let me know if you have any query."