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℃.
When organisms and plants died and sank to the bottom of swamps and oceans, brown soil-like materials called peat are formed. Over millions of years, the peat became covered with sand, clay and other minerals and the peat is converted into layers of sedimentary rocks. After a long time, different type of fossil fuels are formed.
The answer is 4567 grams of ice in the five pound bag to specific heat is the -20 degrees! Hope I helped! :)
Answer:
They are 1.204×10^24 atoms of hydrogen present in 18 grams of water. In order to calculate this,it is necessary to compute the number of hydrogen moles present in the sample.
