2. A student living in a 3-m × 4-m × 4-m dormitory room turns on her 100-W fan before she leaves the room on a summer day, hoping that the room will be cooler when she comes back in the evening. Assuming all the doors and windows are tightly closed and disregarding any heat trans-fer through the walls and the windows, determine the temperature in the room when she comes back 8 h later. Use specific heat values at room temperature and assume the room to be at 100 kPa and 20°C in the morning when she leaves.


Answer 1


The room temperature will be


We know from First law of thermodynamics that the amount of heat flow (Q) through a system is given by

where, 'm' is the mass of the system, is the specific heat and is the temperature difference. Also we know, = 0.718 KJ .

According to the problem, the heat flow can also be written as


Again, if 'm' be the mass of the gas and 'R' be the Universal gas constant = then from Ideal gas equation we can write,


Related Questions

2. A constant voltage is applied to a series RL circuit by closing a switch. The voltage across L is 30 volts at t = 0 and drops to 6 volts at t = .0025 sec. If L = 0.2 H, what must be the value of R?




Resistance R= ?

For RL series circuit

let the voltage across inductance L in the RL circuit be

V₁ = 30 volts at time t₁= 0

the voltage drops to

V₂= 6 volts at time t₂ = 0.0025 sec

the equation to calculate R is

V₂ - V₁ = exp(-(t₂ - t₁ ) R/L)


L ln( V₂ - V₁ )= - (t₂ - t₁)R

R = L ln( V₂ - V₁ )/ (t₁ - t₂)

R= 0.2 x ln(30- 6)/ 0- 0.0025 = -2542.44 Ω

A diver is swimming underneath an oil slick with a thickness of 200 nm and an index of refraction of 1.50. A white light shines straight down towards the diver from above the oil slick. The index of refraction of water is 1.33. What is the longest wavelength of the light in water, ????????????????????????, that is transmitted most easily to the diver?




thickness of oil t = 200 nm

index of refraction μ = 1.5

For transmitted light :---

path difference = 2μ t

For constructive interference

path difference = n λ , λ is wavelength  of light

2μ t = n λ

λ = 2μ t /  n

For longest λ , n = 1

λ = 2μ t

= 2 x 1.5 x 200 nm

= 600 nm

Wavelength in water

= 600 / refractive index of water

= 600 / 1.33

= 451.1 nm Ans

A force of 18 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 14 in. beyond its natural length?







Using Hooke's law to get the spring constant(k)






Work done by spring is given by


Or W=1/2ke²


Work done in stretching the spring to 14in




1 Inch-pounds Force to Joules = 0.113J

Then, to joules



A dog can hear sounds in the range from 15 to 50,000 Hz. What wavelength corresponds to the lower cut-off point of the sounds at 20◦C where the sound speed is 344 m/s?



λ = 22.667 m


The range of frequency (f) of sound wave = 15 Hz to 50 KHz

Temperature (T) = 20 °C

Speed (v) of sound = 344 m/s

Lets assume, wavelength corresponding to the lower cut-off point of the sounds= λ

We know that, the speed (v), frequency (f) and wavelength (λ) are related as


λ = 22.667 m

A force pointing in the xx-direction is given by F=ax3/2F=ax3/2, where aa is a constant. The force does 2.01 kJkJ of work on an object as the object moves from xx = 0 to xxx = 15.2 mm. Find the constant aa.




Given that

F=ax^3/2. a is a constant

The force does a work of

W=2.01KJ from x=0 to x=15.2m

We need to find a

Work is give as,


But this is in x direction only then,

W=∫Fdx. from x=0 to x=15.2m

W=∫ax^3/2dx from x=0 to x=15.2m



W=ax^(2/5)/2.5 from x=0 to x=15.2m

Cross multiply

2.5W=ax^2.5. from x=0 to x=15.2m

2.5W= a (15.2^2.5-0)





A river has a steady speed of 0.360 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point. (b) How much time is required in still water for the same length swim? (c) Intuitively, why does the swim take longer when there is a current?


The given question is incomplete. The complete question is as follows.

A river has a steady speed of 0.360 m/s. A student swims upstream a distance of 1.00 km and swims back to the starting point.

(a) If the student can swim at a speed of 1.29 m/s in still water, how long does the trip take?

(b) How much time is required in still water for the same length swim?

(c) Intuitively, why does the swim take longer when there is a current?


(a)  Relative velocity of the student while going upstream is as follows.

          (1.29 - 0.360) m/s

         = 0.93 m/s

Now, relative velocity of the student while coming downstream is as follows.

          (1.29 + 0.360) m/s

        = 1.65 m/s

When the student is going upstream then time taken to cover 1 km distance is calculated as follows.


         = 1075.26 sec

When the student is going downstream then time taken by it to cover 1 km distance is as follows.


         = 606.06 sec

Hence, total trip time will be calculated as follows.

     Total trip time = (1075.26 sec + 606.06 sec)

                             = 1681.32 sec

(b)  Time taken in still water is calculated as follows.


             = 1550.38 sec

(c)  When there will be current then there will occur a decrease in relative velocity of the swimmer. Therefore, he will take more time.

A small metal ball is given a negative charge, then brought near (i.e., within a few millimeters) to end A of the rod. What happens to end A of the rod when the ball approaches it closely this first time?



They will attract.


We know, in conductors electrons are free to move> in the case given both metal ball and rod are conductor.

Now, when conductor are brought to a charge all the opposite charges moves through the point of conductor which is close to the charge and attraction takes place.

In this case too, when metal ball is bought near rod, rod also develop opposite charge and attraction takes place.  

Hence, this is the required solution.

Charge q1 = 29.5 μC is located at r1 = (3.4 i − 3.5 j) m. A second charge q2 = −44.4 μC is located at r2 = (10.5 i + 7.5 j) m. The force of the second charge on the first charge can be written using unit vectors as follows: F = Fx i + Fy j What is the value of Fx in newtons?



-0.234 N


Electrostatic force is:

Substitute the values:

Negative sign indicates that the force is attractive in nature which is also true as the charges are opposite in nature.

Compute the moon's centripetal acceleration in its orbit around the earth. Recall that the moon orbits the earth every 28.0 ���� and that it is about 240000 ����� from the earth. What force causes this acceleration? Be sure to convert to �� �����.



0.0026 m/s²    


Centripetal acceleration is given as follows:

Substitute the values:

Finding the work done in pulling a stranded climber to safety. In an unfortunate accident, a rock climber finds herself stuck 17 m from the top of a 80 m rock face. Rescuers sent to help the climber lower a harness attached to a cable that will pull the climber to the top of the rock face. If the climber, secured in the harness, weighs 56 kg and the cable weighs 0.7kgm, how much work is done in raising the climber, harness, and cable to the top of the rock face? Note: the weight of the climber and cable is really a mass so you will need to multiply the mass in kg by the acceleration due to gravity (approximately 9.8ms2).



-9446.22 J


Parameters given:

Mass of climber and harness = 56kg

Mass of cable = 0.7kg

Distance between climber and top of rock face = 17m

The work done in pulling the climber is given as:

W = F * d

F is the force applied on the rope. It is opposite the force of gravity pulling the climber, hence, it is given as:

F = -Fg

Fg = force of gravity

Fg = m * g

g = acceleration due to gravity.

The mass of the climber, harness and cable = 56 + 0.7 = 56.7kg

=> Fg = 56.7 * 9.8

Fg = 555.66 N

Therefore, the work done will be:

W = - 555.66 * 17

W = -9446.22 J

The negative value of work means that the work done is opposite the value of the force acting on the climber.

Laptop batteries are rated in W⋅hW⋅h, the product of the power (in WW) that the battery can provide and the time (in hh) that it can provide this power. For instance, a 50 W⋅hW⋅h battery can provide 50 WW for 1.0 hh, 25 WW for 2.0 hh, and so on. A 10.8 VV laptop battery is rated at 74 W⋅hW⋅h.





the charge and the current are related by

Q= I×t

where Q = charge  in columbs

v = 10.8

P = IV

I = P/V = 74/10.8

I =6.85 A

the battery will supply a charge of (6.85×1) C


in 1 hour

how long would it take for a radio wave sent from a space satellite circling mars to reach Earth? Assume that radio waves (a form of electromagnetic radiation) travel at the speed of light?



3 Minutes 2 seconds to 22 minutes 16 seconds


Lets assume the Mars to be at the closest distance to Earth. This distance (D) = 54.6 Million km

The signal travels at the speed of light, so speed of the signal (V) = 300000 km/s

So, the time (T) taken by the radio wave to reach Earth from Mars will be,

Thus, T = 182 Sec = 3 Minutes 2 seconds.

The radio wave will take minimum 03 Minutes 02 seconds to reach Earth. Here is should be noted that the distance between the two planets keep on changing as they revolve around the Sun. There will come a point when Mars is farthest from Earth and the distance (D) will be 401 Million km. Then, the time will change to ,

T = 1336.67 sec = 22.27 Minutes.

So the maximum time will be 22 minutes 16 seconds.

Ice at -50 oF and 1.00 atmis to be melted in a continuous flow process with saturated steam at 1 bar. Find the minimum quantity of steam required per pound of ice.The heat capacity of ice is 0.45 Btu/lbmoF.



The minimum quantity of steam required per pound of ice is -22.5 BTU/lbm


Q = mCT

Q/m = CT

Q/m is the quantity of steam required per pound of ice

C is the heat capacity of ice = 0.45 BTU/lbm°F

T is the temperature of ice = -50 °F

Q/m = 0.45 BTU/lbm°F × -50 °F = -22.5 BTU/lbm

A woman wearing an in-ear hearing aid listens to a television set at a normal volume of approximately60 dB. To hear it, she requires an amplification of30 dB, so the hearing aid supplies sound at90 dB to the ear canal, which we assume to be circular with a diameter of 6.4 mm . What is the output power of the hearing aid?

Express your answer to two significant figures and include the appropriate units.



the output power of the hearing aid in two significant figures is 3.2 × 10⁻⁸ W


Output power of sound is given as;

P = I*A


I is the sound intensity = 90dB = 1 × 10⁻³ W/m²

A is the area through which the transmits;

since the canal is circular in shape.


D as diameter = 6.4mm = 0.0064 m

Output power of the hearing aid = (1 × 10⁻³ W/m²)*(3.217 × 10⁻⁵ m²)

                                                      = 3.2 × 10⁻⁸ W

Therefore, the output power of the hearing aid in two significant figures is 3.2 × 10⁻⁸ W.

Two forces, F? 1 and F? 2, act at a point, as shown in the picture. (Figure 1) F? 1 has a magnitude of 9.20 N and is directed at an angle of ? = 65.0 ? above the negative xaxis in the second quadrant. F? 2 has a magnitudeof 5.80 N and is directed at an angle of ? = 53.9 ? below the negative x axis in the third quadrant. Part A
What is the x component Fx of the resultant force?
Express your answer in newtons.

Part B
What is the y component Fy of the resultant force?
Express your answer in newtons.

Part C
What is the magnitude F of the resultant force?
Express your answer in newtons.

Part D
What is the angle ? that the resultant force forms with the negative x axis? In this problem, assume that positive angles are measured clockwise from the negative x axis.
Express your answer in degrees.



A) x-component of forces is 7.297 N

For x-component FX;

FX = F1 cos 65 + F2 cos 53.9

FX = 9.20 cos 65 + 5.8 cos 53.9

NB both forces are positive if resolved along the x-axis.

FX = 3.88 + 3.417 = 7.297 N

B) 6-component of forces is -3.654 N

For y-component FY

FY = -F1 sin 65 + F2 sin 53. 9

FY = - 9.2 sin 65 + 5.8 sin 53.9

NB F1 is negative along y-axis

FY = - 8.34 + 4.686 = - 3.654 N

C) The magnitude of the resultant force is 8.16 N

Magnitude = (FX^2 + FY^2)^0.5

(7.297^2 + (-3.654)^2)^0.5 = 8.16N

D) Angle made with negative x-axis is 26.56°

Tan^ -1 of FY/FX gives the angle of the resultant force.

= tan^-1 of -3.654/7.297

=tan^-1 of -0.5

= -26.56°

I.e 26.55° from the negative x-axis in a clockwise direction.

If the fraction of vacant sites at a temperature T1 is F1 and the fraction of vacant sites at a temperature T2 is F2, determine the relationship between T1, T2, F1, F2, Qv and k where Qv is the activation energy to form a vacancy and k is Boltzmann’s constant



F₁ / F₂ = exp (-Q_{v} / k)  exp [(T₂-T₁) / T₁T₂]

Q_{v} = -k ΔT / T₁T₂  ln (F₁ / F₂)


The Boltzmann equation for vacancies is

           n = n₀ exp (-Eₙ /kT)

Where n and n₀ are  the number of current and maximum vacancies, respectively, Eₙ is the activation energy (), K the Boltzmann constant and T the absolute temperature


The fraction is

         F = n / n₀ = exp (- Q_{v} / kT)

Let's apply this equation to our case

For temperature T = T₁

       F₁ = exp (-Qv / k T₁)

For temperature T = T₂

      F₂ = exp (-Qv / kT₂)

The ratio of the vacancy fraction is

     F₁ / F₂ = exp (-Q_{v} / k T₁) / exp (- Q_{v} / k T₂)

     F₁ / F₂ = exp (-Q_{v} / k)  exp [(T₂-T₁) / T₁T₂]

In general, the activation energy is sought by clearing this equation

     Q_{v} = -k ΔT / T₁T₂  ln (F₁ / F₂)

An atom of helium and one of argon are singly ionized - one electron is removed from each. The two ions are then accelerated from rest by the electric field between two plates with a potential difference of 150 V. What happens after the two ions accelerate from one plate to the other?

a) The helium ion has more kinetic energy.
b ) The argon ion has more kinetic energy.
c) Both ions have the same kinetic energy.
d) There is not enough information to say which ion has more kinetic energy.


Answer:Both ions have the same kinetic energy.


The kinetic energy of the ions depends on the accelerating potential. The greater the accelerating potential, the greater the kinetic energy of the ions and vice versa. Ions are usually accelerated in such a way that they all have the same kinetic energy. Since the accelerating potential of the He+ and Ar+ are the same, they will both possess the same kinetic energy.

A ball is thrown straight upward. How does the sign of the work done by gravity while the ball is traveling upward compare with the sign of the work done by gravity while the ball is traveling downward?A. Work done by gravity is negative while the ball is traveling upward and positive while the ball is traveling downward.B. Work done by gravity is positive while the ball is traveling both upward and downward.C. Work done by gravity is positive while the ball is traveling upward and negative while the ball is traveling downward.D. Work done by gravity is negative while the ball is traveling both upward and downward.





  • According to the work energy theorem, the work done by external forces on a mass, must equal to the change in kinetic energy.
  • When thrown upward, the change in kinetic energy is negative, due to it must have some initial velocity, which it will diminishing while going upward, as gravity is slowing down it.
  • So, in this part of the trajectory, work done by gravity (only force acting on the ball if we neglect air resistance) must be negative.
  • In the downward part of the trajectory, as the change in kinetic energy is positive (the ball is speeding up when falling), the work done by gravity must be positive too.
  • So, the option A is the one that is true.

The heating coil of a hot water heater has a resistance of 17 Ω and operates at 300 V. How long a time is required to raise the temperature of 115 kg of water from 18°C to 75°C? (The specific heat for water = 10 3 cal/kg ⋅°C and 1.0 cal = 4.186 J).



5183 sec.


  • The power dissipated in the resistance, will be used to raise the temperature of the water completely.
  • This power can be found applying Joule's law, which states the following:


  • If the resistance of the heating coil can be assumed as constant with the temperature, we can find the current I applying Ohm's Law, as follows:


  • Now, from (1) we can find P, as follows:


  • The energy supplied by this power, is just the product of the power times the time during which the energy was delivered:


  • This energy, will be supplied as heat to the mass of water, as stated by the following equation (assuming no heat losses out of the heater):


  • where c= specific heat of water = 4186 J/ºC*kg, m= 115 kg, and ΔT= 57ºC.
  • From (2) and (3), if left sides are equal each other, so do right sides:


  • Replacing by the values, we can solve for  Δt, as follows:


  • The time required to raise the temperature of  115 kg of  water, from 18ºC, to 75ºC, is 5183 sec (approximately 1hr 26').

What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in series? 12210 Correct: Your answer is correct. Ω (b) What is the resistance (in Ω) of thirty-seven 330 Ω resistors connected in parallel?



(a) 12210  Ω

(b) 9.92 Ω


From Electricity,


When resistors are connected in series, the total resistance is the sum of the individual resistance.

From the question,

Rt = R1 + R2 + R3................................. R37 ....................... Equation 1

Where Rt = Total resistance of combined resistance, R1 = resistance of the first resistor, R2 = Resistance of the second resistor .......... and so on up to the 37th resistor.

Note: Since The resistor are identical.

Rt = n(R) ..................... Equation 2

Where n = number of resistor, R = resistance of each resistor.

Given: n = 37, R = 330 Ω

Substitute into equation 2

Rt = 37(330)

Rt = 12210 Ω


For parallel connection,

Rt = R/n...................... Equation 2

Given: R = 330 Ω, n = 37

Substitute into equation 2

Rt = 330/37

Rt = 8.92 Ω

Hence the resistance when connected in parallel = 9.92 Ω

Other Questions
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