Here mass of the iron pan is given as 1 kg
now let say its specific heat capacity is given as "s"
also its temperature rise is given from 20 degree C to 250 degree C
so heat required to change its temperature will be given as
now if we give same amount of heat to another pan of greater specific heat
so let say the specific heat of another pan is s'
now the increase in temperature of another pan will be given as
now we have
now as we know that s' is more than s so the ratio of s and s' will be less than 1
And hence here we can say that change in temperature of second pan will be less than 230 degree C which shows that final temperature of second pan will reach to lower temperature
So correct answer is
<u>A) The second pan would reach a lower temperature.</u>
Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α = 
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹
At sea level, the size amid the 2 alkanes lets for pentane to simmer at a lower temperature than hexane. Phenol has a higher boiling point due to hydrogen bonding High altitude would have the same order while low pressure only cuts the temperature at which a solvent boils. Boiling has to do with molecular size, the occurrence/nonappearance of hydrogen bonds, and other steric issues.
So the answer would be pentane high altitude, hexane high altitude, hexane sea level, hexanol sea level. In order of boil first to boil last. This is clarified because altitude has a better effect on vapor pressure (and hence boiling points) than inter-molecular forces.
0 mark is your answer because you want to start at 0
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where 
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows
![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)
T = 81.52 Deg C
