Of heating. Or when the lake is exposed to boil because of the temperature.
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℃.
Period are going left to right across the periodic table
Groups are going up to down on the periodic table
1. Convert gallons to mL. 1 gal = 3785.4117840007 mL, multiply that by 29 and get 109776.94173602 mL.
2. Since there is one gram per every mL, there are 109776.94173602 g of water in the fish tank.
3. Convert g to pounds. 1 g = 0.0022 pounds. Multiply 109776.94173602 by 0.0022 and end up with about 241.5 pounds of water.
Answer:
Where is the diagram?please put the pictu
