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3 years ago

5

On a hot summer day, a student turns his fan on when he leaves his room in the morning. When he returns in the evening, will the

room be warmer or cooler than the neighboring rooms? Why? Assume all the doors and windows are kept closed.

1 answer:

Virty [35]3 years ago

5 0

Answer:

When the student returns, it will be hotter

Explanation:

We know that the room is closed, so no air is coming in from outside and no air goes outside as the windows and doors are closed. We know the fan is moving and the fan does not cool the air, it only moves the air. Therefore, this process creates kinetic energy that is converted into heat energy (internal) so that room begins to get warmer.

And we know by the the law of Conservation of Energy, there is no loss in energy

So, the room just gets hotter.

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Initially when 1000.00 mL of water at 10oC are poured into a glass cylinder, the height of the water column is 1000.00 mm. The w

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Answer:

$\mathbf{h_2 =1021.9 \ mm}$

Explanation:

Given that :

The initial volume of water $V_1$ = 1000.00 mL = 1000000 mm³

The initial temperature of the water $T_1$ = 10° C

The height of the water column h = 1000.00 mm

The final temperature of the water $T_2$ = 70° C

The coefficient of thermal expansion for the glass is ∝ = $3.8*10^{-6 } mm/mm \ per ^oC$

The objective is to determine the the depth of the water column

In order to do that we will need to determine the volume of the water.

We obtain the data for physical properties of water at standard sea level atmospheric from pressure tables; So:

At temperature $T_1 = 10 ^ 0C$ the density of the water is $\rho = 999.7 \ kg/m^3$

At temperature $T_2 = 70^0 C$ the density of the water is $\rho = 977.8 \ kg/m^3$

The mass of the water is $\rho V = \rho _1 V_1 = \rho _2 V_2$

Thus; we can say $\rho _1 V_1 = \rho _2 V_2$ ;

⇒ $999.7 \ kg/m^3*1000 \ mL = 977.8 \ kg/m^3 *V_2$

$V_2 = \dfrac{999.7 \ kg/m^3*1000 \ mL}{977.8 \ kg/m^3 }$

$V_2 = 1022.40 \ mL$

$v_2 = 1022400 \ mm^3$

Thus, the volume of the water after heating to a required temperature of $70^0C$ is 1022400 mm³

However; taking an integral look at this process; the volume of the water before heating can be deduced by the relation:

$V_1 = A_1 *h_1$

The area of the water before heating is:

$A_1 = \dfrac{V_1}{h_1}$

$A_1 = \dfrac{1000000}{1000}$

$A_1 = 1000 \ mm^2$

The area of the heated water is :

$A_2 = A_1 (1 + \Delta t \alpha )^2$

$A_2 = A_1 (1 + (T_2-T_1) \alpha )^2$

$A_2 = 1000 (1 + (70-10) 3.8*10^{-6} )^2$

$A_2 = 1000.5 \ mm^2$

Finally, the depth of the heated hot water is:

$h_2 = \dfrac{V_2}{A_2}$

$h_2 = \dfrac{1022400}{1000.5}$

$\mathbf{h_2 =1021.9 \ mm}$

Hence the depth of the heated hot water is $\mathbf{h_2 =1021.9 \ mm}$

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Which merchant has the better deal means which merchant offers the farmer a better deal.

For Merchant A, 10 kg bag = $10

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While for Merchant B, 10 kg bag = $8

where the 10kg = 80% dry fertilizer + 20% water content

But the farmer can only use the solid constituents of the bag which means,

Merchant B is giving 80/100 x 10Kg of dry fertilizer for $8

That is, 8kg for $8

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