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3 years ago

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The specific heat of water is 4.184 J g—1 K—1. A piece of iron (Fe) weighing 40 g is heated to 80EC and dropped into 100 g of wa

ter at 25EC. (Specific heat of Fe is 0.446 J g—1 K—1) What is the temperature when thermal equilibrium has been reached

1 answer:

CaHeK987 [17]3 years ago

5 0

Explanation:

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A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t

QveST [7]

Explanation:

(a) First, we will calculate the number of moles as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

= $\frac{\text{100 g}}{4 g/mol}$

= 25 mol

According to the ideal gas equation,

PV = nRT

or, $(P_{2} - P_{1})V = nR (T_{2} - T_{1})$

$(6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T$

$\Delta T$ = 336.17 K

Hence, temperature change will be 336.17 K.

(b) The total amount of heat required for this process will be calculated as follows.

q = $mC \Delta T$

= $100 g \times 5.193 J/g K \times 336.17 K$

= 174573.081 J/K

or, = 174.57 kJ/K (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

3 0

3 years ago

Blood pressure is usually measured by wrapping a closed air-filled jacket equipped with a pressure gage around the upper arm of

Sever21 [200]

Answer:

a) High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

Explanation:

a) <em>High and low pressures in kilopascals</em>:

101.325 kPa equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{high} = 15.999\,kPa$

$p_{low} = 80\,mm\,Hg\times \frac{101.325\,kPa}{760\,mm\,Hg}$

$p_{low} = 10.666\,kPa$

High and low pressures are 15.999 kilopascals and 10.666 kilopascals, respectively.

b) <em>High and low pressures in pounds per square inch</em>:

14.696 psi equals 760 mm Hg, then, we can obtain the values by a single conversion:

$p_{high} = 120\,mm\,Hg\times \frac{14.696\,psi}{760\,mm\,Hg}$

$p_{high} = 2.320\,psi$

$p_{low} = 80\,mm\,Hg\times\frac{14.696\,psi}{760\,mm\,Hg}$

$p_{low} = 1.547\,psi$

High and low pressures are 2.320 pounds per square inch and 1.547 pounds persquare inch, respectively.

c) <em>High and low pressures in meter water column in meters water column</em>:

We can calculate the equivalent water column of a mercury column by the following relation:

$\frac{h_{w}}{h_{Hg}} = \frac{\rho_{Hg}}{\rho_{w}}$

$h_{w} = \frac{\rho_{Hg}}{\rho_{w}}\times h_{Hg}$ (Eq. 1)

Where:

$\rho_{w}$ , $\rho_{Hg}$ - Densities of water and mercury, measured in kilograms per cubic meter.

$h_{w}$ , $h_{Hg}$ - Heights of water and mercury columns, measured in meters.

If we know that $\rho_{w} = 1000\,\frac{kg}{m^{3}}$ , $\rho_{Hg} = 13600\,\frac{kg}{m^{3}}$ , $h_{Hg, high} = 0.120\,m$ and $h_{Hg, low} = 0.080\,m$ , then we get that:

$h_{w, high} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.120\,m$

$h_{w, high} = 1.632\,m$

$h_{w, low} = \frac{13600\,\frac{kg}{m^{3}} }{1000\,\frac{kg}{m^{3}} } \times 0.080\,m$

$h_{w, low} = 1.088\,m$

High and low pressures are 1.632 meters water column and 1.088 meters water column, respectively.

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3 years ago

To determine an epicentral distance scientists consider the arrival times of what wave types

Rudiy27

The answer is P-waves and S-waves

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3 years ago

Choose the word that best completes each sentence.

Igoryamba

Answer:

1.) Tropisms

2.) Thigmotropism

3.) Phototropism

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3 years ago

Read 2 more answers

Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snork

fgiga [73]

Answer:

2354.4 Pa

40221 Pa

Explanation:

$\rho$ = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

$\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa$

The pressure difference in the first case is 2354.4 Pa

$\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa$

The pressure difference in the second case is 40221 Pa

7 0

3 years ago