When the same amount of heat is added to 50.0 g of each substance, which of the following will have the smallest temperature inc
rease?
1 answer:
Answer:
Aluminium
Explanation:
When a certain amount of heat is supplied to a substance, the temperature of the substance increases according to the equation:
where
is the increase in temperature
Q is the amount of heat supplied to the substance
m is the mass of the substance
C is the specific heat capacity of the substance
Here the four substances have the same mass,
m = 50.0 g
Also, the amount of heat supplied to the 4 substances is the same. This means that their increase in temperature is inversely proportional to their specific heat capacity:
Therefore, the substance with the highest specific heat capacity will have the smallest temperature increase: by comparing the values in the table, we see that alluminium has the highest specific heat capacity (0.90 J/gC), therefore it will have the smallest temperature increase.
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Answer:
6.86 m/s
Explanation:
This problem can be solved by doing the total energy balance, i.e:
initial (KE + PE) = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}
Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.
Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.
Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s
The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²
This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)
Now, 1/6m(v0)² = mgL ⇒ v0 =
Hence, v0 = 6.86 m/s