How much heat must be imparted to a piece of ice with a mass of 50 g at 0 ℃ in order to melt it and heat the formed water to a boil?
Initial data: m (mass of ice, water) = 50 g; t (initial ice temperature) = 0 ºС.
Reference data: λ (specific heat of melting of ice) = 340 * 10 ^ 3 J / kg; tboil (final water temperature) = 100 ºС; C (specific heat of water) = 4200 J / (kg * K).
SI system: m = 50 g = 0.05 kg
The amount of heat that must be communicated to a piece of ice: Q = Q1 + Q = λ * m + C * m * (tboil – t) = m * (λ + C * (tboil – t)).
Let’s perform the calculation: Q = 0.05 * (340 * 10 ^ 3 + 4200 * (100 – 0)) = 38,000 J = 38 kJ.
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