Question

Given the following information for an unknown chemical: Specific heat (c) of solid phase: 1.75 J/g°C Heat of fusion (ΔHfus): 32.0 kJ/mol Specific heat (c) of liquid phase: 4.50 J/g°C Heat of vaporization (ΔHvap): 400. kJ/mol Specific heat (c) of vapor phase: 1.50 J/g°C Molar mass: 72.5 g/mol Freezing point: -15.00°C Boiling point: 15.00°C

1.Calculate the total energy needed (in kJ) to convert 15.5g of this chemical from -21.5°C to room temperature (23.0°C).

Answer 1

The answer is 94.8 kJ

At -21.5°C, the chemical is en solid state. Heat required to increase the temperature fom - 21.5°c to its melting point / freezing point, ie, -15.00°C, Q,: Q,= m Csorid AT = 15:59 x 1755 x [-15:00 °C-(-21.5%)] = 15.5g * 1,785 x 6.50€ 176.3125 J Heat required for phase Change from solid to liquid at freezing point, Q₂ : Qq = m Altfus - 15:59 x 320 kJ/mol = 15.59* 32.0 kJ = 6841 kJ 12:59 - Heat required to increase the temperature from -15.00°C to 15.00°c (boiling pront), Q3 : Az = in Chg 4T = 15:59 x 4,50 5x (15:00-(150)]°c ca . 20925 J Scanner

Heat required for phase change from liquid to gas at boiling point, Q4 : Qu= m Attrap = 15:59 x 400.4 I/me = 1550 * x 400 kJ = 85.517 kJ I 7205 Á Heat required to increase the temperature from 15:00°C to 23.0°C, Qg: Qs = n gas AT = 15.59* 150I (230 – 15.0d) & = 1860 J UISAS Now, the total energy required for the Conversion = Q + Q2 +Q₂ + Q4 +Qs = 176.3125 J + 6'8412 J + 2092.5) + 85.517kJ+ 186.05 = (17613125)KI + 6.841KI + 2092-53kJ +85.517 kJ + (18609 kJ = 0. 1763125 kJ + 6.841 kJ + 2:0925 kJ + 85.517 KJ +0.186 kJ ned with = 94.8 kJ Scanner 1000/ 1000/ cs