Question

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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asked 2021-02-19
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..

Answers (1)

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)

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