AP EAMCET20228 Jul 2022Evening ShiftPhysicsThermal Properties of MatterActual
A small electric heater is used to heat 200 ~g of water. The time required to bring all this water from 40^ C to 100^ C is 200 ~s . If specific heat of the water is 4200 ~J ~kg ⁻¹ ~K ⁻¹ then the power supplied by the heater is
Options
- A155 ~W
- B310 ~W
- C88 W
- D252 ~W
Correct answer
D. 252 ~W
Step-by-step solution
Given, t=200 ~s , m=200 ~g =0.2 ~kg aligned s & =4200 jkg ⁻¹ ~K ⁻¹ T & =(100+273)-(40+273)=60 ~K aligned Heat absorbed by water = Energy supplied by heater Power of heater time = mass specific heat 'rise of temperature of water aligned & P t=m s T & P= m s T t & P= 0.2 4200 60 200 aligned =252 ~W