Question

Aluminium ball of mass 0.0 47 kg is heated to a temperature of hundred degree Celsius now this ball is put in a calorimeter of mass 0.14 kg containing water of mass 0.25 kg at 20 degree Celsius the temperature of water rises to 23 degree celsius in thermal equilibrium find out specific heat of aluminium ball

Answer 1

Answer:

Here, mAl=0.047kg ,

T1=100∘C,mcu=0.14kg ,

mw=0.25kg,T0=20∘C

T2=23∘C,sCu=0.386×103Jkg−1K−1

Heat lost by aluminium,

Q1=mAlsCu(T1−T2)

=0.047×sAl×(100−23)

=0.047×sAl×77J

Heat taken by copper calorimeter and water is

Q2=mCusCu(T2−T0)+mwsw(T2−T0)

=0.14×(0.386×103)×(23−20)

+0.25×(4.18×103)(23−20)

=162.12+3135=3197.12J

In the steady state,haet lost= heat gained

∴0.04×sAl×77=3297.12

or, sAl=3297.120.047×77=911Jkg−1K−1.

Explanation: