A liquid of mass 250 g is kept warm in a vessel using an electric heater. The liquid is maintained at `57^(@)` when the power supplied by the heater is 30 W and surrounding temperature is `27^(@)` C . As the heater is switched off, it took 10 s time for the temperature of the liquid to fall from `47^(@)` C to `46.9^(@) C` . The specific heat capacity of the liquid is

Options

A. `8000j kg ^(-1) K^(-1)`
B. `9000j kg ^(-1) K^(-1)`
C. `6000j kg ^(-1) K^(-1)`
D. `12000j kg ^(-1) K^(-1)`

Answer

D

Explanation

Given, power supplied by hearter , `(dQ)/(dt) = 30 W`
mass of liquid , m ` =(250)/(1000) = 250 xx 10^(-3) kg`
`:.` Rate of heat flow, `(dQ)/(dt) = (msdT)/(dt)`
Putting the given values , we get
`rAaa 30 = (250 xx 10^(-3) s xx (47 - 46 . 9))/(10)`
Specific heat capacity of the liquid,
`s = (30 xx10)/(250 xx 10^(-3) xx (0.1)) = 12000 J kg ^(-1) K^(-1)`