Question

Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C, the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 24.1 g of steam at 158°C is condensed, cooled, and frozen to ice at -50.°C.

Answer 1

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Mass of steam = 24.1 8. Molar mass of water = 18.0g/mol oi moles of water = mass Molar mass - 2401 mol to mol i. Heat of vapourisation Hvap TQ2) Total heat released (a) = Heat released by & Steam (158°C -100°c) (Q ) : + Heat released by + water (100°C -0°c) (23) + Heat of busio (Qu) Heat released (0°-50°C) (05) 1 = mcp, DT + AH vapon + m CP₂ DT2 + n al fusion + m By AT3. Here m represents mass & n represents 24•1 2•03 X (158-100) + 2401. x X103 mole + 24.1 * 4.188 (100-0) + 243.4 x 6-02X103 + 24.1 X 4.188 21:1 v 102 103 18 + 2401 x 206 x (0+5950) J => 7781206 J hinn 1708 kJ II ----Scanner ang