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sashaice [31]
3 years ago
9

A 1.00-kg sample of steam at 100.0 °C condenses to water at 100.0 °C. What is the entropy change of the sample if the latent hea

t of vaporization of water is 2.26 x 10⁶ J/kg?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
7 0

Answer:

The entropy change of the sample of water =  6.059 x 10³ J/K.mol

Explanation:

Entropy: Entropy can be defined as the measure of the degree of disorder or randomness of a substance. The S.I unit of Entropy is J/K.mol

Mathematically, entropy is expressed as

ΔS = ΔH/T....................... Equation 1

Where ΔH = heat absorbed or evolved, T = absolute temperature.

<em>Given:  If 1 mole of water = 0.0018 kg,</em>

<em>ΔH = latent heat × mass = 2.26 x 10⁶ × 1 = 2.26x 10⁶ J.</em>

<em>T = 100 °C = (100+273)  K = 373 K.</em>

<em>Substituting these values into equation 1,</em>

<em>ΔS =2.26x 10⁶/373</em>

ΔS = 6.059 x 10³ J/K.mol

Therefore the entropy change of the sample of water =  6.059 x 10³ J/K.mol

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A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o
IgorC [24]

Answer:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

Explanation:

For this case we know the mass of the water given :

m = 787 gr

And we know that the initial temperature for this water is T_i =75 C.

We want to cool this water to the human body temperature T_f = 37 C

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

Q= m c_p \Delta T

Where c_p represent the specific heat for the water and this value from tables we know that c_p =1 \frac{cal}{gr C} for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ

8 0
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Allison wants to determine the density of a bouncing ball. which metric measurements must she use?
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Density depends on mass and volume so option D is correct answer. Hope this helps!
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3 years ago
According to the rule of 72 and about how many years will $78 be worth $39 if the rate of inflation is 5.8%
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The answer is 12.4 years
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series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

Now collect formulas

\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

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v_{ox} = velocity at position "x = 0 "

x = final position

x_{o} = initial position of the object at the start of the motion

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