a .250kg block of aluminum is heated to 100 C and then dropped into .60 kg of water that is initially at 20C. What is the equilibrium temperature of the system? The specific heat of aluminum is .22 kcal/kg C
The sum of the heats gained is zero.
heatgainedAl+heatgainedWater=0
.250*.22*(Tf-00)+.6*c*(Tf-20)=0
solve for Tf
thanks!
To solve this problem, we need to apply the principle of energy conservation.
First, let's calculate the heat absorbed by the aluminum block as it reaches its equilibrium temperature. The equation we'll use is:
Q = m * c * ΔT
Where:
- Q is the heat absorbed (or released) by the object
- m is the mass of the object
- c is the specific heat capacity of the object
- ΔT is the change in temperature of the object
Given values:
- m (mass of aluminum) = 0.250 kg
- c (specific heat of aluminum) = 0.22 kcal/kg C
- ΔT (change in temperature of aluminum) = Tf (final temperature) - Ti (initial temperature)
- Ti (initial temperature of aluminum) = 100 C
Since we don't know the final temperature Tf, we'll leave it as a variable.
Q_aluminum = m * c * ΔT
Q_aluminum = 0.250 kg * 0.22 kcal/kg C * (Tf - 100 C)
Next, let's calculate the heat released by the water as its temperature decreases. The equation to use is the same as above:
Q_water = m * c * ΔT
Given values:
- m (mass of water) = 0.60 kg
- c (specific heat of water) = 1 kcal/kg C
- ΔT (change in temperature of water) = Tf - Ti
- Tf (final temperature of water) = Tf (final temperature)
Q_water = 0.60 kg * 1 kcal/kg C * (Tf - 20 C)
According to the principle of energy conservation, the heat absorbed by the aluminum block is equal to the heat released by the water. Therefore:
Q_aluminum = Q_water
0.250 kg * 0.22 kcal/kg C * (Tf - 100 C) = 0.60 kg * 1 kcal/kg C * (Tf - 20 C)
Now we can solve for Tf, the equilibrium temperature of the system, by solving this equation.