A 100 gram mass of metal at 106°C is placed in 100 grams of cool water at 22°C. The final temperature of the mixture is 25°C. What is the specific heat capacity of the metal?
sum of heats gained is zero
100*cm*(25-106)+200*cw*(25-22)=0
cw is specific heat of water
solve for cm
To find the specific heat capacity of the metal, we need to use the equation:
Q = mcΔT
Where:
Q is the heat transferred
m is the mass of the substance
c is the specific heat capacity
ΔT is the change in temperature
In this case, the heat transferred to the metal is equal to the heat transferred from the metal to the water since no heat is lost to the surroundings. Therefore, we can set up the following equation:
Q gained by metal = Q lost by water
The heat transferred to the metal can be calculated as:
Q gained by metal = mcΔT
The heat transferred from the water can be calculated as:
Q lost by water = mcΔT
Since the final temperature of the mixture is 25°C, the change in temperature for both the metal and water is:
ΔT = final temperature - initial temperature = 25°C - 22°C = 3°C
Using this information, we can solve for the specific heat capacity of the metal.
Let's first calculate the heat transferred to the metal:
Q gained by metal = (mass of metal) * (specific heat capacity of metal) * (change in temperature)
Similarly, let's calculate the heat transferred from the water:
Q lost by water = (mass of water) * (specific heat capacity of water) * (change in temperature)
Since we know the mass and change in temperature for both the metal and water, we can rearrange the equation to find the specific heat capacity of the metal:
Specific heat capacity of metal = (Q gained by metal) / (m * ΔT)
Substituting the known values:
mass of metal = 100 grams
mass of water = 100 grams
change in temperature = 3°C
We need to find the specific heat capacity of the metal. To do that, we need to know the specific heat capacity of water. The specific heat capacity of water is approximately 4.18 J/g·°C.
So, let's calculate the heat transferred to the metal:
Q gained by metal = (100 grams) * (specific heat capacity of metal) * (3°C)
And the heat transferred from the water:
Q lost by water = (100 grams) * (4.18 J/g·°C) * (3°C)
Since Q gained by the metal is equal to Q lost by the water, we can set these two equations equal to each other:
(metal specific heat capacity) * (100 grams) * (3°C) = (100 grams) * (4.18 J/g·°C) * (3°C)
Now we can solve for the specific heat capacity of the metal.