Ask question

Ask question

All categories

4 years ago

13

You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reserv

oir at 0°C, how much heat must you add to it so that both reservoirs have the same increase in entropy?

1 answer:

marysya [2.9K]4 years ago

6 0

Answer:

449.38 J

Explanation:

ΔS = ΔQ/T

Where ΔS = entropy change

Q = quantity of heat

T = temperature

First reservoir :

T = –30°C = - 30 + 273 = 243K

Q = 400 J

Second reservoir :

T = 0°C = 273K

Q =?

To have same increase in entropy for both reservoirs :

Q/T of first reservoir = Q/T of second reservoir

400/243 = Q/273

243 * Q = 400 * 273

Q = (400 * 273) / 243

Q = 109,200 / 243

Q = 449.38271

Q = 449.38 J

You might be interested in

SEPARATION OF VARIABLES - SPHERICAL The potential on the surface of a sphere (radius R) is given by V=V0 cos(2). (Assume V(r=[i

inessss [21]

Answer:

a) Vin(r,θ)=(-Vo/3)+(4Vo/3R^2)r^2P2cosθ

b)σ(θ)=((Voεo)/3R)(20P2cosθ-1)

Explanation:

Due to the complex variables used to solve this exercise, the solution and the explanation are in the image

8 0

3 years ago

A small block with mass 0.0350kg slides in a vertical circle of radius 0.525m on the inside of a circular track. During one of t

kirill115 [55]

Answer:

w = -0.475N

Explanation:

$K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}\\K.E = 0.5mv^{2} \\Normal force at point B, N_{B} = 0.665N\\Normal force at point A, N_{A} = 3.85N\\$

To get Va and Vb

$F = mv_{A} ^{2} /R................(1)\\F = N_{A} - mg.........................(2)\\mv_{A} ^{2} /R = N_{A} - mg\\v_{A} ^{2} = R (N_{A}/m - g)\\v_{A} = \sqrt{ R (N_{A}/m - g)}$

R = 0.525 m

m = 0.0350 kg

g = 9.8 m/s²

$v_{A}= \sqrt{0.525(3.85 /0.0350 - 9.8)} \\v_{A} = 7.25 m/s$

K.Ea = 0.5 * 0.035 * 7.25²

K.Ea = 0.92 J

Since point A is at the bottom of the path, h = 0 m

P.Ea = 0 m

For Vb

$F = mv_{B} ^{2} /R................(1)\\F = N_{B} - mg.........................(2)\\mv_{B} ^{2} /R = N_{B} - mg\\v_{B} ^{2} = R (N_{B}/m - g)\\v_{B} = \sqrt{ R (N_{B}/m - g)}$

$N_{B} = 0.665N$

$v_{B}= \sqrt{0.525(0.665 /0.0350 - 9.8)} \\v_{B} = 2.198 m/s$

$K.E_{B} = 0.5* 0.035 * 2.198^{2} \\K.E_{B} = 0.085 J$

$P.E_{B} = mgh_{B} \\h_{B} = A diameter = 2R = 2 * 0.525\\h_{B} = 1.05 m\\P.E_{B} = 0.035 * 9.8 * 1.05\\P.E_{B} =0.36 J$

$from K.E_{a} + P.E_{a} + W_{fr} = K.E_{b} + P.E_{b}$

$0.92 + w_{fr} + 0 = 0.085 + 0.36\\ w_{fr} = -0.475J$

7 0

4 years ago

A solid uniform sphere and a thin-walled, hollow sphere have the same mass M and radius R. If they roll without slipping up a ra

elixir [45]

Answer:Hollow sphere

Explanation:

Given

same mass for solid and hollow sphere

same $v_{cm}$ before they start up incline

Moment of inertia of solid Sphere

$I_1=\frac{2}{5}Mr^2$

Moment of inertia of hollow sphere

$I_2=\frac{2}{3}Mr^2$

Conserving Energy at bottom and top point for solid sphere

kinetic energy +Rotational Energy=Potential energy

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1$

for pure rolling $v_{cm}=\omega r$

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1$

$\frac{7}{10}Mv_{cm}^2=Mgh_1$

$h_1=\frac{7v_{cm}^2}{10g}$

For hollow sphere

$\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2$

$h_2=\frac{5v_{cm}^2}{6g}$

therefore height gained by hollow sphere is more

6 0

4 years ago

Ahat [919]

<h3>B. True</h3>

"This was the idea that non-living objects can give rise to living organisms."

7 0

3 years ago

PLEASE HELP! I'LL GIVE BRAINLEST

DochEvi [55]

Answer:

1.62 m/s²

Explanation:

4 0

3 years ago