Question

The following information is given for tin at 1 atm: Tb = 2270.00°C AHvap (2270.00°C) = 1.939 x 109 J/g Tm = 232.00°C | AHfus ( 232.00°C) = 59.60 J/g Specific heat solid = 0.2260 J/g °C Specific heat liquid = 0.2430 J/g °C A 22.10 g sample of solid tin is initially at 203.00°C. If 2.822 x 10' J of heat are added to the sample at constant pressure (P = 1 atm), which of the following is/are true? (Select all that apply.) The sample is a solid. The sample is a solid in equilibrium with liquid. The sample is at a temperature of 232.00°C. The sample is a liquid. The sample is at a temperature greater than 232.00°C.

Answer 1

q = m Cp dT T, = 203" as solid = 22.109 x 0.2260 I. (T2-203°c) = 2.822x 1035 - Tz = 768°C This is incorrect as heat due to enthalpy of an fusion is not - included. q=mco d t + malsus it modta = 22.109 x 0.2260 I (232-2039 C + 22-19 *59.61 + 22.109 x 0.24 3001 (T2 - 232)°C = 144.84 +1317.56 + 5.37(T2-232) = 2822 J (12-232) = 253:26 T = 485.26 °C o The Sample is a solid (Initially) E The Sample is a Solid in equilibriom withe liquid (Melting point) ♡ The sample is a liquid (After melting point) ♡ The sample is at a temperature greater than 232°C. 9. Mg(s)-> Mg(s) - Mgles > Mg(e) - 622°C 649° 649% 749° 9= 20.709x 1.047,5 (644-622)€ + 20.90gx 368-35 + 20.700*1-339.(749 – 649)°C = 568.405+ 7623.81 J +2771-73 J = 10,963.94 J = (10.964 KJ]