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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.

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tyaji247

Answer:

The volume of water that was in the kettle is  1170 [tex]cm^{3}[/tex]

Explanation:

Given:

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

Part A:

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

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

i.e. Pt = [tex]m_{s}[/tex][tex]c_{s}[/tex](100 - 20 ) + [tex]m_{w}[/tex][tex]c_{w}[/tex](100 - 20 )

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

[tex]m_{w}[/tex] = 1.17 kg

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

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

∴ v = 1170 [tex]cm^{3}[/tex]

Lanuel

The volume of water in the kettle is equal to 1173  [tex]cm^3[/tex]

Given the following data:

  • Mass of kettle = 710 g
  • Power = 2.0 kW
  • Initial temperature = 20°C
  • Time = 3.5 min

Based on science:

  • Boiling point of water = 100°C
  • Specific heat capacity of water = 4200 kJ/kg°C
  • Specific heat capacity of iron = 450 kJ/kg°C
  • Density of water = 1000 [tex]kg/cm^3[/tex]

To determine the volume of water, in [tex]cm^3[/tex], was in the kettle:

First of all, we would calculate the quantity of energy generated.

[tex]Energy = power \times time\\\\Energy = 2000 \times (3.5 \times 60)\\\\Energy = 2000 \times 210[/tex]

Energy = 420,000 Joules

Next, we would calculate the mass of water by applying the law of conservation of energy:

Total heat energy lost by the electric stove = Total heat energy gained by water and the kettle.

[tex]Q_s = Q_w + Q_k\\\\Q_s = m_wc_w\theta + m_kc_k \theta\\\\m_wc_w\theta = Q_s -m_kc_k\theta\\\\m_w= \frac{Q_s -m_kc_k\theta}{c_w\theta} \\\\m_w = \frac{420,000 - (0.71 \times 450 \times [100-20])}{4200 \times (100-20)}\\\\m_w = \frac{420,000 - 25,560}{336,000} \\\\m_w = \frac{394,440}{336,000} \\\\m_w = 1.173\;kg[/tex]

Mass of water = 1.173 kg.

Now, we can the volume of water by using this formula:

[tex]Density = \frac{Mass}{Volume} \\\\Volume = \frac{Mass}{Density}\\\\Volume = \frac{1.173}{1000}[/tex]

Volume = 0.001173 [tex]m^3[/tex]

In [tex]cm^3[/tex], the volume is:

[tex]Volume = 0.001173 \times 10^6[/tex]

Volume = 1173  [tex]cm^3[/tex]

Read more: https://brainly.com/question/18320053

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