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Heat is supplied to a liquid of mass 500 g contained in a can by passing a current of 4A through a heating coll of resistance 12.5 ohm immersed in the liquid. The initial temperature of the liquid is The liquid reaches its boiling point in 10 minutes after the current Is switched on. It takes a further 2 minutes after the liquid starts to boil for all the liquid to boil away. Calculate: a) the specific heat capacity of the liquid b) the specific latent heat of vaporization of the liquid. (Boiling point of liquid , thermal capacity of the Can=400j/k



Answer :

tanzalice

Let’s break this down step by step:

1.  Initial Information:

  • Mass of the liquid, (m): 500 g (or 0.5 kg)
  • Resistance of the heating coil, (R): 12.5 ohms
  • Current passing through the coil, (I): 4 A
  • Time taken to reach boiling point, (t_1): 10 minutes (or 600 seconds)
  • Additional time after boiling starts, (t_2): 2 minutes (or 120 seconds)
  • Boiling point of the liquid: Not specified
  • Thermal capacity of the can: 400 J/K

2. Energy Supplied:

  • The energy supplied to the liquid is converted into heat.
  • Using Ohm’s law, we find the power dissipated in the heating coil: [ P = I^2R ] [ P = (4 , \text{A})^2 \cdot 12.5 , \Omega = 200 , \text{W} ]

3.  Energy Used for Heating:

    The energy used to raise the temperature of the liquid from its initial

    temperature to the boiling point: [ Q_1 = Pt_1 ]

4.  Energy Used for Vaporization:

    After reaching the boiling point, the remaining energy is used for

    vaporization: [ Q_2 = Pt_2 ]

5.  Total Energy Used:

    The total energy supplied is the sum of (Q_1) and (Q_2): [  

    Q_{\text{total}} = Q_1 + Q_2 ]

6.  Specific Heat Capacity ((c)):

  • The specific heat capacity of the liquid can be calculated using: [ Q_1 = mc\Delta T ] where (\Delta T) is the temperature change from initial to boiling point. [ c = \frac{Q_1}{m\Delta T} ]

7.  Specific Latent Heat of Vaporization ((L)):

   The energy used for vaporization is related to the latent heat: [ Q_2 =

   mL ] [ L = \frac{Q_2}{m} ]

8.  Calculations:

    First, let’s find the total energy:

    -  [ Q_{\text{total}} = Pt_1 + Pt_2 = 200 , \text{W} \cdot 600 , \text{s} +

       200 , \text{W} \cdot 120 , \text{s} = 144000 , \text{J} ]

   -   Next, we need to find the temperature change:

        [ \Delta T = \frac{Q_1}{mc} ] [ \Delta T = \frac{144000 , \text{J}}{(0.5 ,  

        \text{kg})c} Finally, we can calculate (c) and (L).

9.  Results:

  • Specific heat capacity, (c): [calculations]
  • Specific latent heat of vaporization, (L): [calculations]

Please provide the boiling point of the liquid, and I’ll complete the calculations! :)

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