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

e1s2kat26

e1s2kat26

Answered question

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 

Answer & Explanation

Maciej Morrow

Maciej Morrow

Skilled2021-02-20Added 98 answers

What if we add heat Q to a certain object, causing its temperature to rise by the specified amount? T. The following definition of this object's heat capacity, C
C=QT
SI unit : JK=JCo
CA=QT2300
CB=QT2450
2(QT2300)=(QT2450)
Onsolving we get T2=600K

Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-01Added 2605 answers

So, we have

Q=cT

T is the temperature change
The heat that object A transfers per unit mass is represented by;

Q(A)=caT

where;

ca is the item A's specific heat capacity.

According to its mass, the item B transfers the following amount of heat:

Q(B)=cbT

where;

cb is the specific heat capacity of object B

The heat lost by object B is equal to heat gained by object A

Q(A) = -Q(B)

However, item B's heat capacity is two times that of object A.

The two objects' final temperatures are provided by

T2=CaTa+CbTbCa+Cb

T2=CaTa+CbTbCa+Cb

T2=CaTa+2CaTbCa+2Ca

T2=ca(Ta+2Tb)3Ca

T2=Ta+2Tb3

T2=300+(2450)3

T2=400K

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