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:

Answer 1

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.

Answer 2

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

Answer 3

400k

Answer 4

400k

Answer 5

200k

Answer 6

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.

Answer 7

600k

Answer 8

400K

Answer 9

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