AravindG
  • AravindG
REVISION QNS
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
AravindG
  • 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
  • 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
  • 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
  • AravindG
1.kg 2.degree
anonymous
  • anonymous
Degree is not a unit. Rankin, Celsius, Fahrenheit, Kelvin are.
AravindG
  • AravindG
celsius
anonymous
  • 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
  • AravindG
why is it mgh?
anonymous
  • anonymous
Because that is the equation for potential energy, which increases as the boy climbs.
AravindG
  • AravindG
isnt it in*efficiency = mgh?
anonymous
  • anonymous
That would be what we get if the combine the two equations.
AravindG
  • AravindG
ya i gt it!!
AravindG
  • AravindG
k next one
anonymous
  • 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
  • AravindG
k
anonymous
  • 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
  • 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
  • AravindG
srry i did not get the last step
anonymous
  • 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
  • anonymous
Therefore, before mixing the internal energy of the system is expressed as\[U_i = U_{CO_2,i} + U_{O_2,i}\]
AravindG
  • AravindG
k i got that then hw u equated them?
anonymous
  • 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
  • AravindG
k kk cntinu
anonymous
  • anonymous
You can solve that equation for \(T_f\). That's it
anonymous
  • anonymous
Realizing that \(T_i\) is not the same for each species, but \(T_f\) is the same since the gasses are mixed.
AravindG
  • AravindG
k i hava similiar qn tell me if i jst need to apply the above formula for this one too
anonymous
  • 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
  • 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
  • AravindG
shall i use here the same formula above?
anonymous
  • anonymous
Yes, but you need to account for the heat of fusion of the ice.
AravindG
  • AravindG
means ?
AravindG
  • AravindG
an extra term will come?
AravindG
  • AravindG
?
anonymous
  • anonymous
Do you have access to thermodynamic property tables?
AravindG
  • AravindG
no
AravindG
  • AravindG
actually wat u meant by that
anonymous
  • 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
  • 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
  • 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
  • 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
  • anonymous
12 s. End-to-end to me means the same as lengthwise. |dw:1333521257567:dw|
AravindG
  • AravindG
k then?
anonymous
  • anonymous
Do you agree that end-to-end and lengthwise represent the same arrangement?
AravindG
  • AravindG
i think lengthwise is parralel
AravindG
  • AravindG
i maybe wrong
AravindG
  • AravindG
wel the options are a)24 b)3 c)48 d)80 all in celsius
anonymous
  • 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
  • AravindG
srry :P
AravindG
  • AravindG
k...so
anonymous
  • anonymous
If the area increases, what must \(\Delta t\) do such that the equation is still valid?
AravindG
  • AravindG
decrease
anonymous
  • anonymous
How many answers are less than 12?
AravindG
  • AravindG
wow fantastic thx:P
AravindG
  • AravindG
b\y the by how we concluded that area incrases?
anonymous
  • 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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