A liquid of mass 250 g is kept warm in a vessel using an electric heater. The liquid is maintained at 57 ° C 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

Question

A liquid of mass 250 g is kept warm in a vessel using an electric heater. The liquid is maintained at 57 ° C 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, 8000 J kg - 1 K - 1, 9000 J kg - 1 K - 1, 6000 J kg - 1 K - 1, 12000 J kg - 1 K - 1

MARKS App · Accepted Answer

The expression for the rate of heat flow is, d q d t = - k T - T 0 Here T 0 = Surrounding temperature From first condition-| 30 = - k 57 - 27 ⇒ k = - 1 Using second condition- d q d t = - k T - T 0 but d q d t = m s d T d t ∴ m s d T d t = - k T - T 0 Substitute the given values in the above equation. 250 1000 s 47 - 46 . 9 10 = - - 1 47 - 27 ⇒ s = 8000 J kg - 1 K - 1

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