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3 years ago

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For hydrogen gas, H2, tables give values for the parame ters of eqn 7a of a-2728JKmoll, b-3.26x 10 3JK-2 mor and c- 0.50 J K2 mo

l- Calculate the value of the consta pressure heat capacity of hydrogen at 25 °C

1 answer:

Anestetic [448]3 years ago

5 0

Answer:

27.32 J/k Mole.

Explanation:

The constant pressure heat capacity of hydrogen is given by Shomate equation

C_p= a + b×t + c×t^2

C_p = heat capacity (J/mol*K)

t = temperature (K) / 1000.

Temperature =25°C=298 K

t= 298/1000 =0.298

putting values we get

C_p = 27.28+3.26x10^-3 ×0.298+0.50×(0.298)^2

C_p=27.32 J / K mole

the value of the constant pressure heat capacity of hydrogen at 25 °C=

C_p=27.32 J / K mole

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The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. <span>X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.</span>

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4 years ago

Read 2 more answers

suppose 384g of steam originally at 100C is quickly cooled to produce liquid water at 31C. How much heat must be removed from th

dlinn [17]

Answer:

$Q=977216.256\ J=977.216\ kJ$

Explanation:

Given:

mass of steam, $m=384\ g$

temperature of steam, $T_{is}=100^{\circ}C$

temperature of resultant water, $T_{fw}=31^{\circ}C$

We have,

latent heat of vapourization of water, $L=2256\ J.g^{-1}$

specific heat capacity of water, $c=4.186\ J.g^{-1}$

<em>When we cool the steam of 100°C then firstly it loses its latent heat to convert into water of 100°C and the further cools the water.</em>

<u>Now the heat removed from steam to achieve the final state of water:</u>

$\rm Q=latent\ heat\ of\ vapourization+sensible\ heat\ of\ water$

$Q=m(L+c.\Delta T)$

$Q=384(2256+4.186\times (100-31))$

$Q=977216.256\ J=977.216\ kJ$

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3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

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3 years ago

Which of the following solid fuels has the highest heating value?

Makovka662 [10]

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3 years ago

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$

The normal force N is equal to the weight of the car, thus

$F_r=\mu .m.g$

Equating both forces

$\displaystyle \mu .m.g=m.\frac{v^2}{r}$

Simplifying

$\displaystyle \mu =\frac{v^2}{rg}$

Substituting the values

$\displaystyle \mu =\frac{19^2}{(49)(9.8)}$

$\boxed{\mu =0.75}$

7 0

4 years ago