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3 years ago

How much heat is needed to vaporize 10.00 grams of water at 100.0°C? The latent heat of vaporization of water is 2,259 J/g

Physics

2 answers:

Nata [24]3 years ago

8 0

Answer:

Heat of vaporization will be 22.59 j

Explanation:

We have given mass m = 10 gram

And heat of vaporization L = 2.259 J/gram

We have to find the heat required to vaporize 10 gram mass

We know that heat of vaporization is given by $Q=mL$ , here m is mass and L is latent heat of vaporization.

So heat of vaporization Q will be = 10×2.259 = 22.59 J

mina [271]3 years ago

6 0

Answer:

its 22,590 J on edge

Explanation:

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A balloon contains 2.3 mol of helium at 1.0 atm , initially at 240 ∘C. What's the initial volume? What's the volume after the ga

pashok25 [27]

A) initial volume
We can calculate the initial volume of the gas by using the ideal gas law:
$p_i V_i = nRT_i$
where
$p_i=1.0 atm=1.01 \cdot 10^5 Pa$ is the initial pressure of the gas
$V_i$ is the initial volume of the gas
$n=2.3 mol$ is the number of moles
$R=8.31 J/K mol$ is the gas constant
$T_i=240^{\circ}C=513 K$ is the initial temperature of the gas

By re-arranging this equation, we can find $V_i$ :
$V_i = \frac{nRT_i}{p_i} = \frac{(2.3 mol)(8.31 J/mol K)(513 K)}{1.01 \cdot 10^5 Pa}=0.097 m^3$

2) Now the gas cools down to a temperature of
$T_f = 14^{\circ}C=287 K$
while the pressure is kept constant: $p_f = p_i = 1.01 \cdot 10^5 Pa$ , so we can use again the ideal gas law to find the new volume of the gas
$V_f = \frac{nRT_f}{p_f}= \frac{(2.3 mol)(8.31 J/molK)(287 K)}{1.01 \cdot 10^5 Pa} = 0.054 m^3$

3) In a process at constant pressure, the work done by the gas is equal to the product between the pressure and the difference of volume:
$W=p \Delta V= p(V_f -V_i)$
by using the data we found at point 1) and 2), we find
$W=p(V_f -V_i)=(1.01 \cdot 10^5 Pa)(0.054 m^3-0.097 m^3)=-4343 J$
where the negative sign means the work is done by the surrounding on the gas.

5 0

3 years ago

A cell phone is released from the top with the speed of 10ms what is the speed 3s after?

sergeinik [125]

Answer:

30ms

Explanation:

you need to multiple the 10ms by 3s which gives you 30ms

6 0

3 years ago

A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure

Alex787 [66]

V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

   t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

   v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
   speed = |-2| m/s = 2 m/s

Moving right
   V > 0 => t^2 - 9t + 18 > 0
   (t - 6)(t - 3) > 0

   => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)



Going faster and slowing dowm

acceleration, a = v' = 2t - 9
   a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
   Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)

4 0

3 years ago

Speed=100m/sec<br> Frequency=10 Hz<br> Wavelength=?

elena-14-01-66 [18.8K]

Answer:

Wavelength = 10 m

Explanation:

Given:

Speed = 100 m $s^{-1}$

Frequency = 10 Hz = 10 $s^{-1}$

To find : Wavelength = ?

We know that the relationship between wavelength λ, frequency f and speed v is given by the equation

v = fλ

Therefore wavelength λ = v/f

= 100 m $s^{-1}$ / 10 m $s^{-1}$

= 10 m

Hence wavelength = 10 m

4 0

3 years ago

The box leaves position x=0 with speed v0. The box is slowed by a constant frictional force until it comes to rest at position x

Diano4ka-milaya [45]

Answer:

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Explanation:

Given that

At X=0 V=Vo

At X=X1 V=0

As we know that friction force is always try to oppose the motion of an object. It means that it provide acceleration in the negative direction.

We know that

$V^2=U^2-2aS$

$0=V_o^2-2a X_1$

$a=\dfrac{V_o^2}{2X_1}$

So the friction force on the box

Ff= m x a

$Ff=m\times \dfrac{V_o^2}{2X_1}$

Where m is the mass of the box.

4 0

3 years ago