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

`\eta=0.5074\ or\ 50.74\%`

Step-by-step 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* 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* 4.186* 30`

`Q=[tex]\eta=(25116)/(49500)`\ J[/tex]

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

`E=P.t`

`E=1100* 45`

`E=49500\ J`

Now the efficiency:

`\eta=(Q)/(E)`

`\eta=0.5074\ or\ 50.74\%`