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: is C
Explanation:
Both the atomic mass and the atomic number increase from left to right
Answer:
1. not enough dye was added to the drink.
The wrong dye was added to the drink
the water in the drink is evaporating
2. Changing the compound changes the absorbance behavior.
3. Measure the absorbance for the same solution in different cuvette sizes and find the y-intercept.
Explanation:
When the beverage company adds dye to the drink, there should be standard quantity added to the drink so that the color of the drink remains constant. When too much dye is added to the drink, the color will get dark brown or black. When the color of drink get lighter than green this means dye is not added in required quantity.
