Question

REVISION QNS

Answer 1

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?

Answer 2

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

Answer 3

What is the units of the boys mass?

What is the units of the temperature in problem 2?

Answer 4

1.kg
2.degree

Answer 5

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

Answer 6

celsius

Answer 7

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.

Answer 8

why is it mgh?

Answer 9

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

Answer 10

isnt it in*efficiency = mgh?

Answer 11

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

Answer 12

ya i gt it!!

Answer 13

k next one

Answer 14

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.

Answer 15

k

Answer 16

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.

Answer 17

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

Answer 18

srry i did not get the last step

Answer 19

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.

Answer 20

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

Answer 21

k i got that then hw u equated them?

Answer 22

We can write the internal energy of the mixture as\[(mc_v T_f)_{CO_2} + (m c_v T_f)_{O_2}\]

Since, \(U_i = U_{mix}\)
\[(mc_v T_i)_{CO_2} + (mc_v T_i)_{O_2} = (mc_v T_f)_{CO_2} + (mc_v T_f)_{O_2}\]

Answer 23

k kk cntinu

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

shall i use here the same formula above?

Answer 30

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

Answer 31

means
?

Answer 32

an extra term will come?

Answer 33

?

Answer 34

Do you have access to thermodynamic property tables?

Answer 35

no

Answer 36

actually wat u meant by that

Answer 37

They are tables that list thermodynamics properties (enthalpy, internal energy, entropy, specific volume, etc).

Without them, we will have to use the following expression.
\[\left(m \cdot(c_p T_i - h_f)\right)_{ice} + \left(m \cdot c_p T_i \right)_{water} = \left( (m_{ice} + m_{water}) c_p T_f \right)\]

Answer 38

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?

Answer 39

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.

Answer 40

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?

Answer 41

12 s.

End-to-end to me means the same as lengthwise.

Answer 42

k then?

Answer 43

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

Answer 44

i think lengthwise is parralel

Answer 45

i maybe wrong

Answer 46

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

Answer 47

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.

Answer 48

srry :P

Answer 49

k...so

Answer 50

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

Answer 51

decrease

Answer 52

How many answers are less than 12?

Answer 53

wow fantastic thx:P

Answer 54

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

Answer 55

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.