REVISION QNS

- AravindG

REVISION QNS

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- AravindG

1. A loaf of bread gives 5kcal of heat to a boy .What is height he can scale by using this energy ,if his efficiency is 28 percentage and mass is 60?

- AravindG

2.If 22 gm of CO2 at 27 is mixed with 16 gm of O2 at 37 degree .The temperature of the mixture is?

- anonymous

What is the units of the boys mass?
What is the units of the temperature in problem 2?

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## More answers

- AravindG

1.kg
2.degree

- anonymous

Degree is not a unit. Rankin, Celsius, Fahrenheit, Kelvin are.

- AravindG

celsius

- anonymous

For Problem 1.
First, convert kcal to Joules. This conversion factor is given as\[\rm 1 kcal = 4184 J\]
Second, find the amount of energy from the bread the boy can use to climb. This is given as\[\eta = {out \over i n}\] where \(\eta\) is the efficiency, out is the energy the boy can use, and in the energy in the bread.
Finally, from conservation of energy, we can find the height the boy can climb, assuming there is no change in kinetic energy as\[out = mgh\]where out is the energy we solved for in the second part.

- AravindG

why is it mgh?

- anonymous

Because that is the equation for potential energy, which increases as the boy climbs.

- AravindG

isnt it in*efficiency = mgh?

- anonymous

That would be what we get if the combine the two equations.

- AravindG

ya i gt it!!

- AravindG

k next one

- anonymous

We need to make a couple of assumptions.
1) Adiabatic mixing
2) Closed system
3) Constant specific heats (Does not change with temperature)
4) Ideal gas
Realize that\[U_{CO_2} + U_{O_2} = U_{mix}\]where U is the internal energy.

- AravindG

k

- anonymous

Since CO2 and O2 are ideal gases, we can write\[U = c_v(T_f - T_i)\] where \(c_v\) is the constant-volume specific heat.

- anonymous

That should be \[U = m c_v T\]Therefore, we can rewrite the equation as\[\left (m c_v (T_i) \right)_{CO_2} + \left (m c_v ( T_i) \right)_{O_2} = \left (m c_v (T_f) \right)_{mix}\]

- AravindG

srry i did not get the last step

- anonymous

I wrote the equation for internal energy wrong, it should be\[U = m c_v T\]
We are drawing our system boundary such that is encompasses both gasses before they are mixed. After they are mixed, they will occupy the same total volume within the same system boundary. This come from the closed system assumption.

- anonymous

Therefore, before mixing the internal energy of the system is expressed as\[U_i = U_{CO_2,i} + U_{O_2,i}\]

- AravindG

k i got that then hw u equated them?

- anonymous

We can write the internal energy of the mixture as\[(mc_v T_f)_{CO_2} + (m c_v T_f)_{O_2}\]
Since, \(U_i = U_{mix}\)
\[(mc_v T_i)_{CO_2} + (mc_v T_i)_{O_2} = (mc_v T_f)_{CO_2} + (mc_v T_f)_{O_2}\]

- AravindG

k kk cntinu

- anonymous

You can solve that equation for \(T_f\). That's it

- anonymous

Realizing that \(T_i\) is not the same for each species, but \(T_f\) is the same since the gasses are mixed.

- AravindG

k i hava similiar qn tell me if i jst need to apply the above formula for this one too

- anonymous

\[(mc_v T_i)_{CO_2} + (mc_v T_i)_{O_2} = T_f \left((mc_v )_{CO_2} + (mc_v )_{O_2} \right)\]

- AravindG

3.540 gm of ice at 0 celsius is mixed with 540 gm of water at 80 celsius .The final temperature of mixture in celsius will be

- AravindG

shall i use here the same formula above?

- anonymous

Yes, but you need to account for the heat of fusion of the ice.

- AravindG

means
?

- AravindG

an extra term will come?

- AravindG

?

- anonymous

Do you have access to thermodynamic property tables?

- AravindG

no

- AravindG

actually wat u meant by that

- anonymous

They are tables that list thermodynamics properties (enthalpy, internal energy, entropy, specific volume, etc).
Without them, we will have to use the following expression.
\[\left(m \cdot(c_p T_i - h_f)\right)_{ice} + \left(m \cdot c_p T_i \right)_{water} = \left( (m_{ice} + m_{water}) c_p T_f \right)\]

- AravindG

hmm .lemme work out that i will ask if i get a doubt k meanwhile i hav a qn:
is specific heat of a gas in an isothermal process 0?

- anonymous

Absolutely not. Specific heat value doesn't depend on the process by which the fluid is undergoing. Specific heat only depends on the temperature for ideal gasses and temperature and pressure for other fluids.
Isothermal means constant temperature. Iso - same, thermal - temperature.

- AravindG

4.Two rods of same length and material transfer a given amount of heat in 12 s,when they are joined end to end .But when they are joined lengthwise they will transfer same heat in same conditions in how much time?

- anonymous

12 s.
End-to-end to me means the same as lengthwise. |dw:1333521257567:dw|

- AravindG

k then?

- anonymous

Do you agree that end-to-end and lengthwise represent the same arrangement?

- AravindG

i think lengthwise is parralel

- AravindG

i maybe wrong

- AravindG

wel the options are
a)24
b)3
c)48
d)80
all in celsius

- anonymous

They should be in seconds.
We need to know the areas. From Fourier's Law\[{\Delta Q \over \Delta t} = -k A {\Delta T \over \Delta x}\]Assuming Q, k, and the temperature distributions are the same, we need to know how the area changes to find t.

- AravindG

srry :P

- AravindG

k...so

- anonymous

If the area increases, what must \(\Delta t\) do such that the equation is still valid?

- AravindG

decrease

- anonymous

How many answers are less than 12?

- AravindG

wow fantastic thx:P

- AravindG

b\y the by how we concluded that area incrases?

- anonymous

If they are in-fact round rods, the area of contact (mathematically speaking) will be zero.
I'm just assuming that the rods do have a contact area, and that the radius is must smaller than the length.

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