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# The heat capacity of object B is twice that of object A.Initially A is at 300K and B is at 450K. Thy are placed in thermalcontact and the combination

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The heat capacity of object B is twice that of object A.Initially A is at 300K and B is at 450K. Thy are placed in thermalcontact and the combination is isolated. The finaltemperature of both objects is :
A)200K
b)300K
c)400K
d)450K
e)600K
I could use administrator help because these are testquestions I need to correct by tommorrow..Thanks..need explanationof answer..

2021-02-20

Suppose we add the heat Q to a given object, and itstemperature increases by the amount ?T. The heat capacity, C,of this object is defined as follows
$$\displaystyle{C}={\frac{{{Q}}}{{\triangle{T}}}}$$
SI unit : $$\frac{J}{K} = \frac{J}{Co}$$
Fromthe given problem let the heat capacity of the object A be CA,and that of object B is CB
Giventhat heat capacity of object B is twice that of object A
So we can write $$2CA=CB$$
CA=$$\displaystyle{\frac{{{Q}}}{{{T}{2}-{300}}}}$$
CB=$$\displaystyle{\frac{{{Q}}}{{{T}{2}-{450}}}}$$
$$\displaystyle{2}{\left({\frac{{{Q}}}{{{T}{2}-{300}}}}\right)}={\left({\frac{{{Q}}}{{{T}{2}-{450}}}}\right)}$$
Onsolving we get $$T2=600K$$
(So ' e ' is the correct choice)