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:
Heat flux = 13.92 W/m2
Rate of heat transfer throug the 3m x 3m sheet = 125.28 W
The thermal resistance of the 3x3m sheet is 0.0958 K/W
Explanation:
The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:
The heat flux can be defined as the amount of heat flow by unit of area.
Using the previous calculation, we can estimate the heat flux:
It can also be calculated as:
The thermal resistance can be expressed as
For the 3m x 3m sheet, the thermal resistance is
Answer:
P-block metals have classic metal characteristics like they are shiny, they are good conductors of heat and electricity, and they lose electrons easily. These metals have high melting points and readily react with nonmetals to form ionic compounds.
Explanation:
Answer:
O A. A metal higher on the activity series list will replace one that is
lower.
