Ice of 20 g is thrown into the calorimeter containing 100 g of water at a temperature of 20 ° C

Ice of 20 g is thrown into the calorimeter containing 100 g of water at a temperature of 20 ° C at a temperature of -200 ° C. Find the steady-state temperature in the calorimeter. The specific heat capacities of water and ice are 4200 J / kg⋅K and 2100 J / kg⋅K. The specific heat of melting of ice is 332 kJ / kg. Give the answer in degrees Celsius.

Problem data: mw (initial mass of water) = 100 g = 0.1 kg; tv (initial water temperature) = 20 ºС; ml (weight of added ice) = 20 g = 0.02 kg; tl (initial ice temperature) = -20 ºС.

Constants: according to the condition Sv (specific heat capacity of water) = 4.2 * 10 ^ 3 J / (kg * ºС); Сl (specific heat capacity of ice) = 2.1 * 10 ^ 3 J / (kg * ºС); λl (specific heat of melting of ice) = 332 * 10 ^ 3 J / kg.

Heat balance: Sv * mw * (tv – tr) = Sl * ml * (0 – tn) + λl * ml + Sv * ml * (tr – 0).

4.2 * 10 ^ 3 * 0.1 * (20 – tr) = 2.1 * 103 * 0.02 * (0 – (-20)) + 332 * 10 ^ 3 * 0.02 + 4.2 * 10 ^ 3 * 0.02 * (tp – 0).

8.4 – 0.42tr = 0.84 + 6.64 + 0.084tr.

0.504tр = 0.92 and tр = 0.92 / 0.504 = 1.83 ºС.

Answer: The steady-state temperature is 1.83 ºС.




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