A 500 g glass beaker is in thermal equilibrium with 100 ml of 20 Degrees celsius water in it. If a 5 g sample of copper is placed into the water and the system is allowed to come into thermal equilibrium, what will the final temperature of the system be? (assuming no heat loss to the air)
I wonder what kind of glass, and what was the initial temperature of the copper. All that matters.
Sum the heats gained, and set to zero (one will be negative)
heat gained glass+heatgainedwater+heatgainedCu=0
and each of those will be equal to
mass*specificheatformaterial*(Tf-Ti)
is it possible to solve without the initial temperature of the copper?
To find the final temperature of the system, we need to consider the principles of heat transfer and thermal equilibrium.
The equation that governs heat transfer is the specific heat formula:
Q = mcΔT
Where:
Q = Heat energy transferred
m = Mass of the substance
c = Specific heat capacity of the substance
ΔT = Change in temperature
In this scenario, the heat energy lost by the copper sample will be equal to the heat energy gained by the water and the beaker. Since the heat energy lost by the copper is equal to the heat energy gained by the water and the beaker, we can use the equation:
Q(copper) = Q(water + beaker)
Now, let's break down the equation:
For the copper:
Q(copper) = mcΔT
For the water and beaker:
Q(water + beaker) = (mwater + mbeaker)cwaterΔT
We know the following values:
Mass of copper (m) = 5 g
Mass of water (mwater) = 100 g (since the density of water is 1 g/ml, and we have 100 ml of water)
Specific heat capacity of copper (ccopper) = 0.385 J/g°C (provided value)
Specific heat capacity of water (cwater) = 4.184 J/g°C (common value for water)
Initial temperature of the system (Tinitial) = 20°C
Now, let's substitute these values into the equation and solve for the final temperature (Tfinal):
mcΔT = (mwater + mbeaker)cwaterΔT
(5 g)(0.385 J/g°C)(Tinitial - Tfinal) = (100 g + 500 g)(4.184 J/g°C)(Tfinal - Tinitial)
5(0.385)(Tinitial - Tfinal) = 600(4.184)(Tfinal - 20)
1.925(Tinitial - Tfinal) = 2500(Tfinal - 20)
1.925Tinitial - 1.925Tfinal = 2500Tfinal - 50000
1.925Tfinal + 2500Tfinal = 1.925Tinitial + 50000
3425Tfinal = 1.925Tinitial + 50000
Tfinal = (1.925Tinitial + 50000) / 3425
Now, substitute the value of Tinitial into the equation:
Tfinal = (1.925 * 20 + 50000) / 3425
Tfinal ≈ 36.02°C
Therefore, the final temperature of the system would be approximately 36.02°C.