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℃.
Answer:
Macromolecules. A very large organic molecule composed of many smaller molecules, 1)Carbohydrates, 2)proteins, 3)lipids, 4)nucleic acids. Three of the four classes of macromolecules that are polymers. 1.Carbohydrates.
Answer:
Mass of carbon = 109.1 g
Explanation:
Given data:
Mass of carbondioxide = 400 g
Mass of carbon = ?
Solution:
Molar mass of carbon = 12 g/mol
Molar mass of CO₂ = 44 g/mol
Mass of carbon in 400g of CO₂:
Mass of carbon = 12 g/mol/44 g/mol × 400 g
Mass of carbon = 109.1 g
