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3 years ago

7

You make tea with 0.50 kg of 85.0Â°C water and let it cool to room temperature 120.0Â°C2.

1 answer:

dedylja [7]3 years ago

5 0

Answer:

Explanation:

a ) Entropy change dS = dQ/T

= mcdT /T

Integrating both sides

S₂ - S₁ = - mclnT₂ /T₁

= - .5 X 4200 ln (85+273) /( 20 + 273 )

.5 X 4200 ln 358/ 293

= - 417.6 J/K

Entropy change will be negative as heat is lost by the system .

b ) Sine there is no change in the temperature of air , This heat will enter air at temperature ( 20+ 273) K = 293 K

Heat entering air

= .5 x 4200 x 65

= 136500 J

Change in entropy

136500 / 293 ( room temperature is constant at 293k

= + 465.87 J/K

Entropy change will be positive as heat is gained by the system .

Total change in the entropy of the system (tea + air )

= +465.87 - 417.6

= 48.27 J/K

Entropy change will be negative as heat is lost by the system .

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A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

Which body systems help these cells get the energy they need?

PSYCHO15rus [73]

I believe there should be some sort of table attached. Unfortunately I cannot answer this question. Sorry!

5 0

3 years ago

A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward

marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog $\left ( m_d\right )=10kg$

mass of boat $\left ( m_b\right )=46kg$

distance moved by dog relative to ground= $x_d$

distance moved by boat relative to ground= $x_b$

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0= $m_d\times x_d+m_b\dot x_b$

0= $10\times x_d+46\dot x_b$

$x_d=-4.6x_b$

i.e. boat will move opposite to the direction of dog

Now

$|x_d|+|x_b|=7.8$

substituting $x_d$ value

$5.6|x_b|=7.8$

$|x_b|=1.392m$

$|x_d|=6.4032m$

now the dog is 22.5-6.403=16.096m from shore

4 0

3 years ago

A net force of 1.0N acts on a 4.0 kg object, initially at rest, for 4.0 seconds. What is the distance the object moves during th

Eva8 [605]

Use F=ma to solve for a=0.25m/s^2. Now to get distance you need to integrate this twice to get
x=(1/2)at^2. Plug in a and t to get
x=2m

8 0

3 years ago

"Suppose you tie a rock to the end of a 0.96 m long string and spin it in a horizontal circle with a constant angular velocity o

Ber [7]

Answer:

The mass of the rock is $m = 2.46406 \ kg$

Explanation:

From the question we are told that

The length of the string is $l = 0.96 \ m$

The angular velocity is $w = 20.25 \ rad/s$

The tension on the string is $T = 970 \ N$

Generally the centripetal force acting on the rock is mathematically evaluated as

$T = mlw^2$

making m the subject of the formula

$m = \frac{T}{w^2 * l}$

substituting values

$m = \frac{970}{(20.25^2) * (0.96)}$

$m = 2.46406 \ kg$

8 0

3 years ago