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99 Percentile Qs Bank for JEE MainPhysicsKinetic Theory of Gases

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 liq

Options

  1. A8000   J   kg - 1   K - 1
  2. B9000   J   kg - 1   K - 1
  3. C6000   J   kg - 1   K - 1
  4. D12000   J   kg - 1   K - 1

Correct answer

A. 8000   J   kg - 1   K - 1

Step-by-step solution

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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