Answer: The molar heat capacity of aluminum is
Explanation:
As we know that,
.................(1)
where,
q = heat absorbed or released
= mass of water = 130.0 g
= mass of aluminiunm = 23.5 g
= final temperature
=
= temperature of water =
= temperature of aluminium =
= specific heat of water=
= specific heat of aluminium= ?
Now put all the given values in equation (1), we get
Molar mass of Aluminium = 27 g/mol
Thus molar heat capacity =
Answer:
Explanation:
Applied force, F = 18 N
Coefficient of static friction, μs = 0.4
Coefficient of kinetic friction, μs = 0.3
θ = 27°
Let N be the normal reaction of the wall acting on the block and m be the mass of block.
Resolve the components of force F.
As the block is in the horizontal equilibrium, so
F Cos 27° = N
N = 18 Cos 27° = 16.04 N
As the block does not slide so it means that the syatic friction force acting on the block balances the downwards forces acting on the block .
The force of static friction is μs x N = 0.4 x 16.04 = 6.42 N .... (1)
The vertically downward force acting on the block is mg - F Sin 27°
= mg - 18 Sin 27° = mg - 8.172 ... (2)
Now by equating the forces from equation (1) and (2), we get
mg - 8.172 = 6.42
mg = 14.592
m x 9.8 = 14.592
m = 1.49 kg
Thus, the mass of block is 1.5 kg.
