A 100 g mass of aluminum at 100 deg C is placed in 100 g of water at 10 deg C. The final temperature of the mixture is 25 deg C. What is the specific heat of the aluminum?
mahesh sawant
To calculate the specific heat of aluminum, we can use the formula:
Q = mcΔT
Where:
Q = heat transferred
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
In this case, we know the following values:
m (mass of aluminum) = 100 g
m (mass of water) = 100 g
ΔT (change in temperature) = 25°C - 10°C = 15°C
Since the heat transferred to the aluminum is equal to the heat transferred from the water, we can write:
Q (aluminum) = Q (water)
Using the equation Q = mcΔT, we can rewrite this as:
m (aluminum) * c (aluminum) * ΔT (aluminum) = m (water) * c (water) * ΔT (water)
Substituting the known values:
100 g * c (aluminum) * 15°C = 100 g * c (water) * 15°C
Notice that the mass and change in temperature are the same for both substances, so we can cancel them out:
c (aluminum) = c (water)
Therefore, to find the specific heat of aluminum, we need to know the specific heat of water. The specific heat capacity of water is approximately 4.186 J/g°C.
Thus, the specific heat of aluminum can be approximated as 4.186 J/g°C.
To determine the specific heat of aluminum, we can use the equation:
Q = mcΔT
Where:
Q = heat transferred
m = mass of the substance
c = specific heat capacity
ΔT = change in temperature
In this scenario, the heat transferred from the aluminum to the water is equal to the heat gained by the water. Since no heat is lost to the surroundings, we can assume that:
Q(aluminum) = Q(water)
First, let's calculate the heat transferred from the aluminum to the water. We can use the equation:
Q = mcΔT
For the aluminum:
m(aluminum) = 100 g
c(aluminum) = specific heat of aluminum (unknown)
ΔT(aluminum) = final temperature − initial temperature
= 25°C − 100°C
Now, the heat transferred from the aluminum to the water is equal to the heat gained by the water. Let's calculate the heat gained by the water:
Q(water) = mcΔT
For the water:
m(water) = 100 g
c(water) = specific heat of water (known) = 4.18 J/g°C (approximate value)
ΔT(water) = final temperature − initial temperature
= 25°C − 10°C
Since Q(aluminum) = Q(water), we can equate the two equations:
m(aluminum) * c(aluminum) * ΔT(aluminum) = m(water) * c(water) * ΔT(water)
Plugging in the values:
100 g * c(aluminum) * (25°C − 100°C) = 100 g * 4.18 J/g°C * (25°C − 10°C)
Now solve for c(aluminum):
c(aluminum) = (100 g * 4.18 J/g°C * (25°C − 10°C)) / (100 g * (25°C − 100°C))
c(aluminum) = (418 J * 15°C) / (−7500 J)
c(aluminum) = −6270 J / −7500 J
c(aluminum) ≈ 0.836 J/g°C
Therefore, the specific heat of aluminum is approximately 0.836 J/g°C.
q1-q2=0
q1=q2
q1=mc1∆t
q2=mc2∆t
Masses are the same so they aren't needed.
∆t=-75 C°, and c1=???? -75 will just be 75.
∆t=15 C°, and c2=4.186 J/g• C°
c1*(75 C)=(15 C)*(4.186 J/g• C°)
75c=62.79
c=62.79/75
c=0.8372 J/g• C°
Someone check my thinking.