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Answer:
25.2°C
Explanation:
Given parameters:
Energy applied to the water = 1000J
Mass of water = 50g
Final temperature = 30°C
Unknown:
Initial temperature = ?
Solution:
To solve this problem, we use the expression below:
H = m c Ф
H is the energy absorbed
m is the mass
c is the specific heat capacity
Ф is the change in temperature
1000 = 50 x 4.184 x (30 - initial temperature )
1000 = 209.2(30 - initial temperature)
4.78 = 30 - initial temperature
4.78 - 30 = - initial temperature
Initial temperature = 25.2°C
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℃.
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