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:

ans is 400 K how?

I got it now.

(2c)(450-T) = (c)(T-300)
Solve for T

Yes, assuming they are the same mass.

To find the final temperature, we can use the principle of heat transfer. When two objects of different temperatures are in thermal contact and there is no heat exchange with the surroundings, they will reach a final temperature, which is a common equilibrium temperature.

Let's first calculate the amount of heat gained or lost by each object. The heat gained or lost by an object can be calculated using the formula:

Q = mcΔT

where Q is the heat gained or lost, m is the mass of the object, c is its specific heat capacity, and ΔT is the change in temperature.

Given that the heat capacity of object B is twice that of object A, we can say that the specific heat capacity of B is twice that of A, i.e., cB = 2cA.

For object A:
Q1 = mA * cA * ΔTA

For object B:
Q2 = mB * cB * ΔTB

Since the combination is isolated, the total heat lost by object A must be equal to the total heat gained by object B, so Q1 = Q2.

mA * cA * ΔTA = mB * cB * ΔTB

Dividing both sides by mA * cA, we get:
ΔTA = (mB / mA) * (cB / cA) * ΔTB
= 1 * (2cA / cA) * ΔTB
= 2ΔTB

Now, substitute the values given:
ΔTA = 2ΔTB
300 K - TfinalA = 2(TfinalB - 450 K)

Simplifying, we get:
-150 K = 2TfinalB - 900 K
2TfinalB = 750 K

Divide both sides by 2:
TfinalB = 375 K

Since the final temperature of both objects is the same, we have:
TfinalA = TfinalB = 375 K

Therefore, the correct answer is 375 K, not 400 K.

To determine the final temperature of objects A and B when they are placed in thermal contact and isolated, we can use the principle of conservation of energy.

The heat lost by object B will be equal to the heat gained by object A. This can be expressed as:

Q_lost = Q_gained

The heat lost by object B can be calculated using the formula:

Q_lost = C_B * ΔT_B

Where C_B is the heat capacity of object B and ΔT_B is the change in temperature of object B.

Similarly, the heat gained by object A can be calculated as:

Q_gained = C_A * ΔT_A

Where C_A is the heat capacity of object A and ΔT_A is the change in temperature of object A.

Given that the heat capacity of object B is twice that of object A, we can write:

C_B = 2 * C_A

Substituting the values into the equation, we get:

C_A * ΔT_A = 2 * C_A * ΔT_B

Since the objects are in thermal contact and isolated, the heat lost by object B is equal to the heat gained by object A. This implies:

ΔT_B = -ΔT_A

Substituting this into the equation, we have:

C_A * ΔT_A = 2 * C_A * (-ΔT_A)

Simplifying the equation, we get:

ΔT_A = -2 * ΔT_A

Dividing both sides by ΔT_A, we find:

-1 = -2

This implies that the change in temperature of object A is equal to -1, which means the final temperature of object A will be 299 K (300 K - 1 K).

Since ΔT_A = -ΔT_B, the final temperature of object B will be 450 K - 299 K = 151 K.

Therefore, the final temperature of both objects A and B when placed in thermal contact and isolated is 299 K for object A and 151 K for object B, not 400 K as mentioned in your question.