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nataly862011 [7]

2 years ago

6

During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one c

ycle occurs every 10 s, how many minutes will it take to freeze 570 g of water, initially at 0°C?(Lv = 80 cal/g

1 answer:

Alexandra [31]2 years ago

6 0

Answer:

138.18 minutes

Explanation:

$L_v$ = Latent heat of water at 0°C = 80 cal/g

m = Mass of water = 570 g

Heat removed for freezing

$Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal$

Let N be the number of cycles and each cycle removes 56 cal from the freezer.

So,

$55\times N=45600\\\Rightarrow N=\frac{45600}{55}$

Each cycle takes 10 seconds so the total time would be

$\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes$

The total time taken to freeze 138.18 minutes

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According to Kepler’s Third Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

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For satellite Y, the orbital period $T_{Y}$ is:

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