Ask question

Ask question

All categories

3 years ago

11

Why does sand heat faster than water

2 answers:

Levart [38]3 years ago

8 0

Answer: sand should both heat and cool faster than water

Explanation: water has a higher specific heat capacity than sand- meaning that it takes a lot of heat, or energy, to raise the temperature of water one degree, whereas it takes comparatively little energy to change the temperature of sand by one degree

snow_lady [41]3 years ago

5 0

The sand should both heat and cool faster than the water. This is because water has a higher specific heat ca- pacity than sand – meaning that it takes a lot of heat, or energy, to raise the temperature of water one degree, whereas it takes comparatively little energy to change the temperature of sand by one degree.

You might be interested in

A mixture of A and B is capable of being ignited only if the mole percent of A is 6 %. A mixture containing 9.0 mole% A in B flo

atroni [7]

Explanation:

As it is given that mixture (contains 9 mol % A and 91% B) and it is flowing at a rate of 800 kg/h.

Hence, calculate the molecular weight of the mixture as follows.

Weight = $0.09 \times 16.04 + 0.91 \times 29$

= 27.8336 g/mol

And, molar flow rate of air and mixture is calculated as follows.

$\frac{800}{27.8336}$

= 28.74 kmol/hr

Now, applying component balance as follows.

$0.09 \times 28.74 + 0 \times F_{B} = 0.06F_{p}$

$F_{p}$ = 43.11 kmol/hr

$F_{A} + F_{B} = F_{p}$

$F_{B}$ = 43.11 - 28.74

= 14.37 kmol/hr

So, mass flow rate of pure (B), is $F_{B}$ = $14.37 \times 29$

= 416.73 kg/hr

According to the product stream, 6 mol% A and 94 mol% B is there.

Molecular weight of product stream = $Mol. weight \times 43.11 kmol/hr$

= $0.06 \times 16.04 + 0.94 \times 29$

= 28.22 g/mol

Mass of product stream = 1216.67 kg/hr

Hence, mole of $O_{2}$ into the product stream is as follows.

$0.21 \times 0.94 \times 43.11$

= $8.5099 kmol/hr \times 329 g/mol$

= 272.31 kg/hr

Therefore, calculate the mass % of $O_{2}$ into the stream as follows.

$\frac{272.31}{1216.67} \times 100$

= 22.38%

Thus, we can conclude that the required flow rate of B is 272.31 kg/hr and the percent by mass of $O_{2}$ in the product gas is 22.38%.

4 0

3 years ago

Water Erosion: Cause and Effect

Cloud [144]

Answer:

Definitely add more points

Explanation:

I'm saying this because this is worth way more points

3 0

3 years ago

Inside a room, the air is often warmer near the ceiling then near the floor. Which of the following accounts for this difference

Vanyuwa [196]

The correct answer is C. convection

4 0

3 years ago

Juan observes a material in the rocks of a hilluan hammers off a piece and then examines the pieces with a hana tena He

Nata [24]

Answer:

I definitely think he mostly observed that it was clear in color.

Explanation:

5 0

3 years ago

A sample has a mass of 7.412 grams and a volume of 0.217 liters. What is the density of this sample?

Sever21 [200]

Answer : The density of a sample is, 34.2 g/L

Explanation :

Density : It is defined as the mass contained per unit volume.

Formula used for density :

$Density=\frac{Mass}{Volume}$

Given :

Mass of a sample = 7.412 grams

Volume =0.217 L

Now put all the given values in the above formula, we get the density of a sample.

$Density=\frac{7.412g}{0.217L}=34.2g/L$

Therefore, the density of a sample is, 34.2 g/L

3 0

3 years ago