Question

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?

Answer 1

Answer:

$\eta=0.5074\ or\ 50.74\%$

Explanation:

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

GIVEN:

• density $\rho=1\ g.mL^{-1}$

• specific heat $c=4.186\ J.g^{-1}.K^{-1}$

• mass of coffee, $m=200\times 1=200\ g$

• initial temperature of coffee, $T_i=30^{\circ}C$

• final temperature of coffee, $T_f=60^{\circ}C$

• power rating of oven, $P=1100\ W$

• time taken to reach the final temperature, $t=35\ s$

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

$Q=m.c.\Delta T$

$Q=200\times 4.186\times 30$

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

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

$E=P.t$

$E=1100\times 45$

$E=49500\ J$

Now the efficiency:

$\eta=\frac{Q}{E}$

$\eta=0.5074\ or\ 50.74\%$