The heat capacity of object B is twice that of object A. Initially A is at 300 K and B is at 450 K. They are placed in thermal contact and the combination is isolated. The final temperature of both objects is:
Let the heat capacity of A = c. The heat capacity of B is 2c. Let the final temperature be T.
The heat lost by B when A and Be are placed together is: (2c)(450-T)
The heat gained by A is (c)(T-300).
Assuming the Law of conservation of Energy holds, and there are no heat losses (heat transferred elsewhere),
(2c)(450-T) = (c)(T-300)
Solve for T.
400k
The basic relationships above are:
1. Heat transferred = (heat capacity)(temperature change), or
deltaH = C•(T2-T1)
( T1, T2 = initial and final temperatures )
2. Heat lost by B = Heat gained by A
400k
600k
Qb=mb.cb.(T-450) Qa=ma.ca.(T-300)
Cb=2Ca
Qlost=Qgain
mbcb(T-450)=maca(T-300)
Cb.(T-450)= Ca.(T-300)
2Ca.(T-450)=Ca.(T-300)
2(T-450)=(T-300)
2T-900=T-300
T=600k
400K
Its 400K but i cant quite figure it out numerically. You can guess it by general thinking that if B loses temp and A gains then the resultant temp at which thermal equilibrium is achieved will be in between their initial temp.
To find the final temperature of both objects, we can use the principle of heat transfer. When two objects are in thermal contact and isolated, they tend to reach thermal equilibrium, where their temperatures equalize.
In this case, we need to consider the heat exchanged between the two objects until they reach equilibrium. The heat exchanged between two objects is given by the equation:
Q = m * c * ΔT
Where:
Q is the heat exchanged
m is the mass of the object
c is the heat capacity
ΔT is the change in temperature
Given that the heat capacity of object B is twice that of object A, we can write it as:
cB = 2 * cA
Now, let's consider the heat exchanged by object A and B until they reach equilibrium. The heat exchanged can be assumed to be equal for both objects as they are in thermal contact and isolated. We can write:
mA * cA * ΔTA = mB * cB * ΔTB
If we divide both sides of the equation by ΔTA, we get:
mA * cA = mB * cB * (ΔTB/ΔTA)
Now, let's plug in the given values:
mA * cA = mB * 2 * cA * (ΔTB/ΔTA)
Now, let's simplify the equation:
mA = 2 * mB * (ΔTB/ΔTA)
We know that ΔTA is the change in temperature of object A, which is final temperature - initial temperature:
ΔTA = Tf - 300 K
Similarly, ΔTB is the change in temperature of object B, which is final temperature - initial temperature:
ΔTB = Tf - 450 K
Now, let's substitute these values back into the equation:
mA = 2 * mB * ((Tf - 450 K)/(Tf - 300 K))
Since the mass of object B is not given, we can assume it to be 1 (unit mass). Therefore, the equation becomes:
mA = 2 * (Tf - 450 K)/(Tf - 300 K)
Now, let's solve this equation to find the value of final temperature (Tf).
First, let's cross multiply:
mA * (Tf - 300 K) = 2 * (Tf - 450 K)
Expand the equation:
mA * Tf - 300 K * mA = 2 * Tf - 900 K
Rearrange the equation:
(mA - 2) * Tf = 300 K * mA - 900 K
Tf = (300 K * mA - 900 K)/(mA - 2)
Now, plug in the values of mA and solve the equation to find Tf.