Thermodynamics and power??

Question

The burner on an electric stove has a power output of 2 kW . A 660 g stainless steel tea kettle is filled with 20 degree C water and placed on the already hot burner. If it takes 2.60 min for the water to reach a boil, what volume of water, in cm^3, was in the kettle? Stainless steel is mostly iron, so you can assume its specific heat is that of iron.

Answer 1

Use the relationship

dQ = mcdT for each material Since dQ is change in energy you need to consider time rat of change:

dQ/dt = mc dT/dt

Now for the water plus kettle:

dQ/dt = 2 kW =(mk*cfe + mw*cw)*dT/dt

mk = mass of kettle, mw = mass of water, cfe = specific heat of iron, and cw= specific heat of water.

Now mw = pw*V where pw=density of water = 1 gm/cm^3 , and V = volume of water so;

dQ/dt = 2 kW =(mk*cfe + pw*V*cw)*dT/dt

Now dT/dt = (100 - 20)/(2.6 min*60 sec/min) = 0.513 K/sec

So everything but V is know - solve for V

V = [dQ/dt/(dT/dt) - mk*cfe]/(pw*cw)