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℃.
The heat transfer formula is;
Q = m * c * Δ T >>>> (1)
where, Q is the heat transfer
m = mass (gram)
c = the specific heat capacity (J/g)
Δ T = change in temperature
∵ we have one mole of Ethanol
∴ the weight of ethanol equals its molecular weight = (2*12)+(6*1)+(16) = 46 g
we will assume that the specific heat capacity of ethanol is 2.46 J/g (from google)
ΔT = 25 - 320 = - 295 C
By substitution in (1)
∴ Q = 2.46 * 46 * (-295) = - 33382.2 J
Answer:
it gets rust and could not run properly
Answer: 4.5 x 10e-7
Explanation: 450 x 1e+9 = correct answer
Multiply amount of nanometers by 1e+9 to get the approximate result in meters.
