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Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. Your microwave oven draws 1100 W of electrical power when it is running. If it takes 45 s for this microwave oven to raise the temperature of the coffee to 60∘C , what is the efficiency of heating with this oven?

Respuesta :

Answer:

[tex]\eta=0.5074\ or\ 50.74\%[/tex]

Explanation:

Considering the density & specific heat capacity of coffee to be equal to that of water.

GIVEN:

  • density [tex]\rho=1\ g.mL^{-1}[/tex]
  • specific heat [tex]c=4.186\ J.g^{-1}.K^{-1}[/tex]
  • mass of coffee, [tex]m=200\times 1=200\ g[/tex]
  • initial temperature of coffee, [tex]T_i=30^{\circ}C[/tex]
  • final temperature of coffee, [tex]T_f=60^{\circ}C[/tex]
  • power rating of oven, [tex]P=1100\ W[/tex]
  • time taken to reach the final temperature, [tex]t=35\ s[/tex]

Heat released by the coffee to come to 60°C:

[tex]Q=m.c.\Delta T[/tex]

[tex]Q=200\times 4.186\times 30[/tex]

[tex]Q=[tex]\eta=\frac{25116}{49500}[/tex]\ J[/tex]

Now the energy used by the oven in the given time:

[tex]E=P.t[/tex]

[tex]E=1100\times 45[/tex]

[tex]E=49500\ J[/tex]

Now the efficiency:

[tex]\eta=\frac{Q}{E}[/tex]

[tex]\eta=0.5074\ or\ 50.74\%[/tex]

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