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:
C. Atomic mass
Explanation:
i just did the quiz and got 100 :) i hope I get brainiest answer
This one is the easiest law, but you would take 53 and 185 and add them together to get 235 and then you will minus 235 and 365 and the answer you are looking for is 130 mmHg! Hopefully this helped you!!
Answer:
It is because water molecules in the air condensed on to the container of the drink.
Explanation:
The way this works is the water molecules outside are hot and in the gas state, so when they come into contact with the cold side of the container they lose energy due to heat transfer between the molecules and the container, becoming a liquid on the side of the drink.
Answer:
Ea=5.5 Kcal/mole
Explanation:
Let rate constant are 
and 
at temperature 
and 
By using Arrhenius equation at two different two different temperature,
By putting value of R=2 cal/mole.K
By rounding off upto 2 significant figure;
