An insulated chamber of 700 kg is divided into two equal halves as shown in figure 8 below. Initially, the left hand side (A) contains 0 oC ice and the right hand side (B) contain 45 oC water.Find the final temperature of the system at equilibrium.
I don't see the diagram/picture/attachment. It is difficult for me to picture what you've described. Don't you need to know how much ice and how much water is in A and B? Is that 350 kg ice in side A and 350 kg water in side B?
45
To find the final temperature of the system at equilibrium, we need to apply the principle of conservation of energy.
The principle of conservation of energy states that energy cannot be created or destroyed; it can only be transferred or transformed from one form to another.
In this case, the heat energy will flow from the higher temperature (45 oC water) to the lower temperature (0 oC ice) until the system reaches equilibrium.
To calculate the final temperature, we need to determine the amount of heat transferred and use it to find the equilibrium temperature.
To do this, we can use the equation:
Q = mcΔT
where:
Q is the amount of heat transferred
m is the mass
c is the specific heat capacity
ΔT is the change in temperature
First, we need to calculate the heat transferred from the water to the ice. Since the chamber is divided into two equal halves, each half contains 350 kg of the substance. Let's assume the specific heat capacity of water is 4.18 J/g°C and the specific heat capacity of ice is 2.09 J/g°C.
The heat transferred from the water to the ice can be calculated as follows:
Qwater = (mwater)(cwater)(ΔTwater)
Qwater = (350 kg)(4.18 J/g°C)(0 - 45°C)
Qwater = -322,350 J
The negative sign indicates that heat is being transferred from the water to the ice.
Next, we calculate the heat transferred from the ice to the water:
Qice = (mice)(cice)(ΔTice)
Qice = (350 kg)(2.09 J/g°C)(0 - (-45)°C)
Qice = -161,175 J
Now, we can find the total heat transferred by summing up the heat transferred from the water to the ice and from the ice to the water:
Qtotal = Qwater + Qice
Qtotal = -322,350 J + (-161,175 J)
Qtotal = -483,525 J
Finally, we divide the total heat transferred by the total mass of the system to find the change in temperature:
ΔTtotal = Qtotal / (mwater + mice)
ΔTtotal = -483,525 J / (700 kg + 700 kg)
ΔTtotal = -483,525 J / 1400 kg
ΔTtotal = -345.375°C
The negative sign indicates that the system's final temperature is below absolute zero. However, the concept of absolute zero is not physically possible, so we assume the system has reached its lowest possible temperature.
Therefore, the final temperature of the system at equilibrium is 0°C.