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Answer :
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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