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A liquid of density 0 . 85 g cm - 3 flows through a calorimeter at the rate of 8 . 0 cm 3 s - 1 . Heat is added by means of a 250 W electric heating coil and a temperature difference of 15 ° C is established in steady-state conditions between the inflow and the outflow points of the liquid. The specific heat for the liquid will be ( 1   kcal   = 4186   J )

Options

  1. A0 . 6   kcal   kg - 1 K - 1
  2. B0 . 3   kcal   kg - 1 K - 1
  3. C0 . 5   kcal   kg - 1 K - 1
  4. D0 . 4   kcal   kg - 1 K - 1

Correct answer

A. 0 . 6   kcal   kg - 1 K - 1

Step-by-step solution

This is a problem on 'flow calorimeter' used to measure the specific heat of the liquid. Amount of heat supplied to the water per second by the heating coil = Q s = 250   J = 2 5 0 4 1 8 6 kcal The volume of liquid flowing out per second = 8 . 0   cm 3   =   8   ×   10 - 6   m 3 Mass of this liquid = ( 0 . 85 )   ×   1000   ×   8   ×   10 - 6   kg The temperature rise of this mass of liquid = 15 °   C Hence,

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