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kobusy [5.1K]
3 years ago
11

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r

eservoir 2 is at a temperature of 354 K. Compute the total change in entropy arising from the conduction of 1760 J of heat through the wire. Please give your answer in units of J/K.
Physics
1 answer:
Schach [20]3 years ago
8 0

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= \frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}

= \frac{-1760}{563} + \frac{1760}{354}

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

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Answer:

false statement : b )  For the motion of a cart on an incline plane having a coefficient of kinetic friction of 0.5, the magnitude of the change in kinetic energy equals the magnitude of the change in gravitational potential energy

Explanation:

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Read 2 more answers
An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider compl
pickupchik [31]

Answer:

a. 2.1 s

b.0.48 Hz

c. A=24cm

d. 72cm/s

Explanation:

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

What are the  period,

period is the time taken for a wave particle to make one complete oscillation

a) 31 / 15 = 2.066 seconds

= 2.1 s

(b) frequency : this the number of oscillation made in one seconds.

it is also the inverse of the period.

= oscillations / time

= 15/31= 0.48 Hz

(c) amplitude : maximum displacement from the origin

amplitude = 1/2 of the difference of oscillation marks

= 1/2(57-10) = 47/2cm

23.5cm

A=24cm

(d) maximum speed of the glider?

V=ωA

angular frequency *Amplitude

V=a*pi*f*amplitude

2π x frequency x amplitude = maximum speed

= 2π x .48 x 24

=72.38 cm/s

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