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 : The final equilibrium temperature of the water and iron is, 537.12 K
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.
where,
= specific heat of iron = 560 J/(kg.K)
= specific heat of water = 4186 J/(kg.K)
= mass of iron = 825 g
= mass of water = 40 g
= final temperature of water and iron = ?
= initial temperature of iron = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:
Therefore, the final equilibrium temperature of the water and iron is, 537.12 K
Answer:
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This statement is True! Lets think about it... When water boils, the water doesnt evaporate from the bottom of the bowl it evaporates from the top!
=)
