Dr. Lievers loves fresh fruit salad, but he is extremely particular about the wa

Question

Dr. Lievers loves fresh fruit salad, but he is extremely particular about the way in which it is prepared and eaten. For example, he believes that fruit salad should be stored at exactly 5 degree C and he has a small refrigerator that he uses exclusively for that purpose. This refrigerator uses 300 W of power and has a coefficient of performance of 1.7. One day Dr. Lievers goes to the store, buys some fresh fruit, brings it home, washes it, cuts it up into exact 1.5cm cubes and mixes it together in his preferred proportions: 1 kg of peaches, 1 kg of watermelon, 1kg of honeydew melon. He also adds 0.5 kg of (uncubed!) blueberries. If the fruit is initially at room temperature (20 degree C), how long would the refrigerator take (in minutes) to cool his fruit salad to 5 degree C? If he was willing to wait up to 30 minutes, what is the maximum mass (in kg) of fruit salad he could prepare and cool to 5 degree C? Use the specific heats given below. Ignore any energy needed to cool the container holding the fruit salad.

Explanation / Answer

We know that

COP of a refrigerator= Heat removal rate/Power supplied

1.7=Heat removal rate/300

So, Heat removal rate of the given refrigerator =1.7*300=510 J/s

(a) Total heat to be removed to cool the Salad from 20C to 5C= Heat removed by all ingredients of salad (Bluberries+Honey dew melon+peaches+watermelon)

=(0.5*3.64+1*3.92+1*3.81+1*3.94)*(20-5)=202.35 kJ=202350J

So time taken to cool the salad from 20C to 5C=202350/510=396.76s=6.61minutes

(b) From above we have seen that 3.5 kg of salad(0.5 kg Bluberries,1 kg Honeydew melon,1 kg peaches & 1kg watermelon) takes 6.61 minute to cool from 20C to 5C ,

So, in 30 minute the maximum mass of salad which we can prapare =3.5*30/6.61=15.88 kg of salad

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