Question

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and placed on the already hot burner. Part A If it takes 3.5 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

Answer:

The volume of water that was in the kettle is  1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of  $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$, then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$$c_{s}$(100 - 20 ) + $m_{w}$$c_{w}$(100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$