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A certain mass of ice at -20 degrees C is mixed with 2.5 litres of water at 20 degrees C. What mass of ice is needed so that the final equilibrium temperature of the mixture is exactly 0 degrees C? Given: rho_water = 1002.5 kg/m³, latent heat of fusion of ice L = 80 cal/g, specific heat of ice S_ice = 0.5 cal/(g*deg C), specific heat of water S_water = 1 cal/(g*deg C).
- 1 kg
- 5 kg
- 20 kg
- All of the above are correct
Correct answer: All of the above are correct
Solution
Heat lost by 2.5 L water = 2506.25 * 20 ≈ 50125 cal. For final T = 0 deg C, any amount of ice m_ice such that (i) m_ice is large enough that not all of it melts, i.e., Q_w <= Q₁ + m_ice*80, and (ii) large enough to absorb all the water's heat. Both 1 kg and 5 kg satisfy these constraints, making all listed options valid (within simplifying assumptions).
Related JEE Advanced Physics questions
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