The calorimeter contains water with a mass of mw = 500 g at a temperature of t0 = 20 ° C. How much steam (mp), having a temperature tc = 100 ° C, must be admitted into the calorimeter to raise the temperature to t = 40 ° C? The boiling point of water tк = 100 oC, the specific heat of vaporization r = 2.3 10 ^ 6 J / kg, the specific heat capacity of water sv = 4200 J / (kg K).
Given: mw (mass of water) = 500 g (0.5 kg); tv (water temperature) = 20 ºС; tp (water vapor temperature) = 100 ºС; tр (equilibrium temperature) = 40 ºС.
Constants: L (beats heat of vaporization) = 2.3 * 10 ^ 6 J / kg; Cw (specific heat capacity of water) = 4200 J / (kg * ºС).
The mass of water vapor is determined from the equality: Sv * mw * (tp – tw) = mp (L + Sv * (tp – tp)), whence mp = Sv * mw * (tp – tw) / (L + Sv * (tp – tp)).
Calculation: mp = 4200 * 0.5 * (40 – 20) / (2.3 * 10 ^ 6 + 4200 * (100 – 40)) = 0.0165 kg (16.5 g).
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