Q26 / 160 · NTA Abhyas (JEE Main) 2020 · Physics · Thermodynamics

A cooking vessel on a slow burner contains 5 kg of water and an unknown mass of ice in equilibrium at 0oC at time t = 0. The temperature of the mixture is measured at various times and the result is plotted as shown in diagram. During the first 50 min the mixture remains at 0oC. From 50 min to 60 min, the temperature increases to 2oC. Neglecting the heat capacity of the heat capacity of the vessel, the initial mass of the ice is

Figure

  • A
    1 0 7 kg
  • Correct answer
    B
    5 7 kg
  • C
    5 4 kg
  • D
    5 8 kg

Solution

Let m be the mass of ice.
Rate of heat given by the burner is constant.
In the first 50 min

           dQ dt = mL t 1 = m kg × 8 0 × 4.2 × 1 0 3 J/kg 5 0  min                ...(i)

From 50 min to 60 min

     dQ dt = m + 5 S H 2 O Δ θ t 2

          = m + 5 4.2 × 1 0 3 J/kg × 2 ∘ C 1  min                ...(ii)

From Eqs. (i) and (ii)

      8 0 m 5 0 = 2 m + 5 1 0

      7 m = 5 ⇒ m = 5 7 kg ≃ 0.7 kg

Hence, the correct answer is option: 'B'.

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