Answer:
The specific heat of the metal is 2.09899 J/g℃.
Explanation:
Given,
For Metal sample,
mass = 13 grams
T = 73°C
For Water sample,
mass = 60 grams
T = 22°C.
When the metal sample and water sample are mixed,
The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.
Since, heat lost by metal is equal to the heat gained by water,
Qlost = Qgain
However,
Q = (mass) (ΔT) (Cp)
(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)
After mixing both samples, their temperature changes to 27°C.
It implies that
, water sample temperature changed from 22°C to 27°C and metal sample temperature changed from 73°C to 27°C.
Since, Specific heat of water = 4.184 J/g°C
Let Cp be the specific heat of the metal.
Substituting values,
(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)
By solving, we get Cp =
Therefore, specific heat of the metal sample is 2.09899 J/g℃.
Molality
is one way of expressing concentration of a solute in a solution. It is expressed
as the mole of solute per kilogram of the solvent. To calculate for the
molality of the given solution, we need to convert the mass of solute into
moles and divide it to the mass of the solvent.
<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>
<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality
= 0.7543 mol / kg</span>
<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is
0.7453 molal.</span>
Answer:
The amount of water converted from liquid to gas with 6,768 joules is approximately 3.035 g
Explanation:
The amount of heat required to convert a given amount of liquid to gas at its boiling point is known as the latent heat of evaporation of the liquid
The latent heat of evaporation of water, Δ
≈ 2,230 J/g
The relationship between the heat supplied, 'Q', and the amount of water in grams, 'm', evaporated is given as follows
Q = m × Δ
Therefore, the amount of water, 'm', converted from liquid to gas at the boiling point temperature (100°C), when Q = 6,768 Joules, is given as follows;
6,768 J = m × 2,230 J/g
∴ m = 6,768 J /(2,230 J/g) ≈ 3.035 g
The amount of water converted from liquid to gas with 6,768 joules = m ≈ 3.035 g.
