If we takes 11. 2 kj of energy to raise the temperature of 145 g of benzene from 23. 0°c to 68. 0°c, then the specific heat of benzene is
1716.475 J/kg.°C.
Specific Heat capacity is given as
Q = cm(t₂-t₁) ................
Where Q = quantity of heat, c = specific heat of benzene, m = mass of benzene, t₁ = initial temperature, t₂ = final temperature.
=
Given:
Substitute into equation 2.
Hence the specific heat of benzene =
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Answer:
6.25 grams is the mass of solute dissolved.
Explanation:
w/w % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.
Mass of the solution = 50.0 g
Mass of the solvent = x
w/w % = 12.5%
x = 6.25 g
6.25 grams is the mass of solute dissolved.
Explanation:
As density is defined as the mass of a substance divided by its volume.
Mathematically, Density =
It is given that mass is 50 g and density is 0.934 .
Hence, calculate the volume of methyl acetate as follows.
Density =
0.934 =
Volume =
or, = (as 1
= 1 mL)
Thus, we can conclude that the volume of methyl acetate the student should pour out is .
According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.
So, at equilibrium, =
,
+
q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature
=(15X13X4.19)+(148X88.3X4.19)
= 81.37 ° C