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JulsSmile [24]
3 years ago
13

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 p

laced 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.
Physics
1 answer:
asambeis [7]3 years ago
4 0

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}

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