In an aluminum pan weighing 500 g, in which there is 0.5 l of water and 200 g of snow at 0 ° C, 150 g of water

| April 19, 2021 | education

In an aluminum pan weighing 500 g, in which there is 0.5 l of water and 200 g of snow at 0 ° C, 150 g of water vapor is admitted at a temperature of 120 ° C. Taking the heat capacity of steam equal to 2 kJ / (kg * ˚С), find the temperature established in the system.

To determine the value of the temperature established in the specified system, we will use the equality: Cp * mp * (tp – tkond) + L * mp + Sv * mp * (tkond – tp) = Ca * mk * (tp – t0) + Sv * mw * (tр – t0) + λl * msn + Sv * msn * (tр – t0).

Constants and variables: Cn – beats. heat capacity of superheated steam (according to the condition Cn = 2 * 10 ^ 3 J / (kg * ºС)); mp is the mass of water vapor (mp = 0.15 kg); tp – initial steam temperature (tp = 120 ºС); tkond – condensation temperature (tkond = 100 ºС); L – beats. heat of condensation (L = 2.3 * 10 ^ 6 J / kg); Cv – beats heat capacity of water (Cw = 4.2 * 10 ^ 3 J / (kg * ºС)); Ca – beats heat capacity of aluminum (Ca = 920 J / (kg * ºС)); mk is the mass of the pan (mk = 0.5 kg); t0 is the initial temperature of the pan with snow and water (t0 = 0 ºС); mw is the initial mass of water in the pan (mw = 0.5 kg (corresponding to V = 0.5 l)); λl – beats the heat of melting of ice (λl = 3.4 * 10 ^ 5 J / kg); msn – initial mass of snow (msn = 0.2 kg);.

Calculation: 2 * 10 ^ 3 * 0.15 * (120 – 100) + 2.3 * 10 ^ 6 * 0.15 + 4.2 * 10 ^ 3 * 0.15 * (100 – tr) = 920 * 0.5 * (tr – 0) + 4.2 * 10 ^ 3 * 0.5 * (tr – 0) + 3.4 * 10 ^ 5 * 0.2 + 4.2 * 10 ^ 3 * 0, 2 * (tр – 0).

2 * 10 ^ 3 * 0.15 * (120 – 100) + 2.3 * 10 ^ 6 * 0.15 + 4.2 * 10 ^ 3 * 0.15 * (100 – tr) = 920 * 0, 5 * (tr – 0) + 4.2 * 10 ^ 3 * 0.5 * (tr – 0) + 3.4 * 10 ^ 5 * 0.2 + 4.2 * 10 ^ 3 * 0.2 * (tр – 0).

6 * 10 ^ 3 + 345 * 10 ^ 3 + 63 * 10 ^ 3 – 630tr = 460tr + 2100tr + 68 * 10 ^ 3 + 840tr.

4030tr = 346 * 10 ^ 3 and tp = 346 * 10 ^ 3/4030 = 85.86 ºС.



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