Question

a electric heater of power 1000 raises the temperature of a 5kg of liquid from 25 to 31 degrees in 2 minutes
Calculate:
1.The heat capacity
2.the specific heat capacity of the liquid

Answer 1

As Heat is:
Q=mc∆t
=>Pxt=mc∆t
=>1000x2=5xcx6
=>c=200/3 J/K.

Answer 2

w= mct
p=w/t
w=p*t
p*t=mct
1000*120=5×c×7
120000/5*6=c
4000 J/kg/k
heat capacity = p*t/initial temperature
=1000*120/6
=120000/6
=20000J/k
I hope it will help you!!!!!