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Nookie1986 [14]
3 years ago
12

What minimum heat is needed to bring 250 g of water at 20 ∘C to the boiling point and completely boil it away? The specific heat

of water is 4190 J/(kg⋅K) and its heat of vaporization is 22.6×105 J/kg.
Physics
1 answer:
Amiraneli [1.4K]3 years ago
4 0

Answer:633.8 KJ

Explanation:

Given

mass of water\left ( m\right )=250gm

Initial temperature\left ( T_i\right )=20^{\circ}C

Final temperature \left ( T_f\right )=100^{\circ}C

Specific heat of water \left ( c \right )=4190 J/kg-k

heat of vaporization\left ( L\right )=22.6\times 10^5 J/kg

Heat required for process\left ( Q\right )=heat to raise water temperature from 20 to 100 +Heat to vapourize water completely

Q=mc\left ( T_f-T_i\right )+mL

Q=0.25\times 4190\times \left ( 100-20\right )+0.25\times 22\times 10^5

Q=\left ( 0.838+5.5\right )\times 10^5

Q=6.338\times 10^5J=633.8 KJ

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A flute is designed so that it plays a frequency of 261.6 Hz, middle C, when all the holes are covered and the temperature is 20
irinina [24]

Answer:

0.655 m

13.468°C

Explanation:

v = Speed of sound at 20.0°C = 343 m/s (general value)

For one both end open we have the expression

L=\dfrac{\lambda_1}{2}\\\Rightarrow L=\dfrac{\dfrac{v}{f_1}}{2}\\\Rightarrow L=\dfrac{\dfrac{343}{261.6}}{2}\\\Rightarrow L=0.655581039755\ m

The length of the flute is 0.655 m

Beat frequency is given by

\Delta f=f_1-f_2\\\Rightarrow 3=261.6-f_2\\\Rightarrow f_2=261.6-3\\\Rightarrow f_2=258.6\ Hz

Velocity of the wave is

v=f_2\lambda_1\\\Rightarrow v=258.6\times \dfrac{343}{261.6}\\\Rightarrow v=339.066513761\ m/s

The temperature is given by

T=273(\dfrac{v}{331})^2\\\Rightarrow T=273(\dfrac{339.066513761}{331})^2\\\Rightarrow T=286.468227799\ K=286.468227799-273=13.468227799^{\circ}C

The temperature of the room is 13.468°C

4 0
3 years ago
A force of 3,200 kg x m/s^2 (Newton’s) acts on a truck giving it an acceleration of 2 m/s^2. What is the mass of the truck ?
Ivan

Answer:

According to newton's second law of motionF=ma Data:-F=3200kgm/sec² or N ,a=2m/sec² ,m=? solution :-F=ma here we have to find m so m=F/a ,m=3200/2=1600kg

3 0
3 years ago
The peak wavelength of light coming from a giant red star is 660 nm. Calculate the approximate surface temperature of this star
laiz [17]

Answer:

The value is T =260.33 \ F

Explanation:

From the question we are told that  

  The the peak wavelength is  \lambda_p =  660 nm = 660 *10^{-9} \  m

 Generally according to the Wien's displacement law

       \lambda_p *  T =  2.898*10^{-3} \ m \cdot K  

Here T is the approximate surface temperature of this star in K so

       660*10^{-9} *  T =  2.898*10^{-3} \ m \cdot K  

=>    T =  4091 \  K

Converting to Fahrenheit ,

     T = [400 - 273.15 ] * \frac{9}{5} + 32

=>  T =260.33 \ F

5 0
2 years ago
Question 6 Unsaved How many neutrons does element X have if its atomic number is 29 and its mass number is 84? Your Answer:
belka [17]

84-29=55 neutrons in nucleus


3 0
3 years ago
What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?
VikaD [51]

(a) 1.33\cdot 10^{-11} m

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

e \Delta V = \frac{hc}{\lambda}

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

e=1.6\cdot 10^{-19}C is the electron's charge

\Delta V = 93.3 kV = 93300 V is the potential difference

h=6.63\cdot 10^{-34} Js is the Planck constant

c=3.00\cdot 10^8 m/s is the speed of light

\lambda is the wavelength of the emitted photon

Solving the formula for \lambda, we find:

\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

E=\frac{hc}{\lambda}

where

h is Planck constant

c is the speed of light

\lambda=1.33\cdot 10^{-11} m is the wavelength of the photon, calculated previously

Substituting,

E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

E=\frac{1.50\cdot 10^{-14} J}{1.6\cdot 10^{-19} J/eV}=93300 eV

(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

e \Delta V = \frac{hc}{\lambda}

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

3 0
3 years ago
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