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## Question

You bring home a fresh whole watermelon from your local fruit stand. At sale time it weighted exactly 10kg. Once at home you use a “Laser Densiometer” to measure the watermelon’s water density. (a Laser Densiometer, a machine of my invention, allows you to measure any object’s water density. You zapp with the Densiometer and the machines reads the density of the object in percentage. For example, a “ 25%” reading is an object that has 25% water content) When just brought home, the watermelon’s density reads 99%. You leave the watermelon untouched in your kitchen. After a week you use the Densiometer once again on the watermelon and it now reads 98%. It must have been hot in the kitchen and some water must have evaporated from the untouched watermelon!

How much does the watermelon weight now?

## 6 answers to this question

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You bring home a fresh whole watermelon from your local fruit stand. At sale time it weighted exactly 10kg. Once at home you use a "Laser Densiometer" to measure the watermelon's water density. (a Laser Densiometer, a machine of my invention, allows you to measure any object's water density. You zapp with the Densiometer and the machines reads the density of the object in percentage. For example, a " 25%" reading is an object that has 25% water content) When just brought home, the watermelon's density reads 99%. You leave the watermelon untouched in your kitchen. After a week you use the Densiometer once again on the watermelon and it now reads 98%. It must have been hot in the kitchen and some water must have evaporated from the untouched watermelon!

How much does the watermelon weight now?

Assuming the densiometer reading means that the watermelon was 99% water by weight, then originally, the watermelon had 9.9 kg water and 0.1 kg other stuff. One week later, it presumably still has the 0.1 kg other stuff, but now should have only 4.9 kg water (0.1 / 0.02 = 4.9). The watermelon now weighs only 5.0 kg - eww, shriveled up watermelon...

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another way to solve this is

out of 10Kg 0.1 kg was mass rest water.

Now after some time, 2% of wt = 0.1 Kg (as it read 98%density)

So wt = 5Kg

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You bring home a fresh whole watermelon from your local fruit stand. At sale time it weighted exactly 10kg. Once at home you use a “Laser Densiometer” to measure the watermelon’s water density. (a Laser Densiometer, a machine of my invention, allows you to measure any object’s water density. You zapp with the Densiometer and the machines reads the density of the object in percentage. For example, a “ 25%” reading is an object that has 25% water content) When just brought home, the watermelon’s density reads 99%. You leave the watermelon untouched in your kitchen. After a week you use the Densiometer once again on the watermelon and it now reads 98%. It must have been hot in the kitchen and some water must have evaporated from the untouched watermelon!

How much does the watermelon weight now?

Initially, dry mass:total mass = .01

After a week, dry mass:total mass = .02

Since dry mass is a constant, total mass must have reduced by a factor of two, meaning the watermelon weighed 5 kg after one week.

That's a shriveled melon main.

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density = mass/volume.

Initially, the watermelon weighed 10 kg. After some time, the water content evaporated from the melon and the volume of the melon reduces.

So, now mass = volume * density.

mass = constant.

Both the volume and the density change to keep the mass constant.

So, still, after some water content has evaporated, the mass of the melon remains 10kg.

Note: In this case, the dry mass remains a constant and the volume of the melon changes.

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Initially, the watermelon weighed 10 kg. After some time, the water content evaporated from the melon and the volume of the melon reduces.

So, now mass = volume * density.

mass = constant.

Both the volume and the density change to keep the mass constant.

mass = constant? Where did you get that from? You're merely asserting that and not showing why that's true.

In your solution you state "water content evaporated" and then state "Note: In this case, the dry mass remains a constant."

The mass of the watermelon did not remain constant. Since dry mass has remained constant and water has been removed via evaporation, concluding that the watermelon's weight is unchanged is an incorrect one.

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mass = constant? Where did you get that from? You're merely asserting that and not showing why that's true.

In your solution you state "water content evaporated" and then state "Note: In this case, the dry mass remains a constant."

The mass of the watermelon did not remain constant. Since dry mass has remained constant and water has been removed via evaporation, concluding that the watermelon's weight is unchanged is an incorrect one.

ohh yeah. I was thinking about density and writing about mass. I realise it now.

My theory here is wrong....

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