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 :&nbsp;A)200K&nbsp;b)300K&nbsp;c)400K&nbsp;d)450K&nbsp;e)600K&nbsp;

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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 :&amp;nbsp;A)200K&amp;nbsp;b)300K&amp;nbsp;c)400K&amp;nbsp;d)450K&amp;nbsp;e)600K&amp;nbsp;

Maciej Morrow · Accepted Answer

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&#039;s heat capacity, CC=Q△TSI unit :&amp;nbsp;JK=JCoCA=QT2−300CB=QT2−4502(QT2−300)=(QT2−450)Onsolving we get&amp;nbsp;T2=600K

Jeffrey Jordon · Accepted Answer

So, we haveQ=c△T△T&amp;nbsp;is the temperature changeThe heat that object A transfers per unit mass is represented by;Q(A)=ca△Twhere;ca is the item A&#039;s specific heat capacity.According to its mass, the item B transfers the following amount of heat:Q(B)=cb△Twhere;cb is the specific heat capacity of object BThe heat lost by object B is equal to heat gained by object AQ(A) = -Q(B)However, item B&#039;s heat capacity is two times that of object A.The two objects&#039; final temperatures are provided byT2=CaTa+CbTbCa+CbT2=CaTa+CbTbCa+CbT2=CaTa+2CaTbCa+2CaT2=ca(Ta+2Tb)3CaT2=Ta+2Tb3T2=300+(2⋅450)3T2=400K

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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