Thermochemistry of mixing quantities of ice and steam:

Transcription

1 Thermochemistry of mixing quantities of ice and steam: Consider example problem number 4, from Lect-11: ex. 4 When 40.0 g of steam at 125 deg.c are mixed with 125 g of ice at 15 deg.c, what is the final state and final temperature of the mixture? Steam, and ice, are single phases of the pure substance water. Gain / loss of heat in any single phase is readily calculated by: q = ( sp. ht. )( mass)( T) ( phase) However, single phases only exist in restricted temperature ranges. When water is subjected to an external pressure of one atmosphere, then... SOLID WATER can only exist at temperatures of ZERO degrees C OR LOWER, LIQUID WATER can only exist at temperatures between ZERO and 100 degrees C, and GAS WATER can only exist at temperatures of 100 degrees C OR HIGHER. Specific temperatures of ZERO and 100 degrees C are normal freezing and boiling point temperatures for water. TWO PHASES are present at these stages, and gain / loss of heat causes removal of one phase while temperature remains constant.

2 Compare the thermochemistry of ONE PHASE vs. TWO: Thermochemistry of a SINGLE PHASE: using water as an example... H 2 O (solid) specific heat = 2.03 J / g-deg gain/loss of heat = q = ( 2.03 J / g-deg)x(mass)x( T ) when subjected to an external pressure of one atmosphere, H 2 O (solid) CAN ONLY EXIST AT TEMPERATURES OF ZERO DEGREES CELSIUS OR LOWER. H 2 O (liquid) specific heat = 4.18 J / g-deg gain/loss of heat = q = ( 4.18 J / g-deg)x(mass)x( T ) when subjected to an external pressure of one atmosphere, H 2 O (liquid) CAN ONLY EXIST AT TEMPERATURES BETWEEN ZERO AND 100 DEGREES CELSIUS. H 2 O (gas) specific heat = 2.02 J / g-deg gain/loss of heat = q = ( 2.02 J / g-deg)x(mass)x( T ) when subjected to an external pressure of one atmosphere, H 2 O (gas) CAN ONLY EXIST AT TEMPERATURES OF 100 DEGREES CELSIUS AND HIGHER.

3 Thermochemistry FOR TWO PHASES OF A PURE COMPONENT: At PHASE-CHANGE temperatures, TWO PHASES are present and temperature is fixed. Gain / loss of heat causes removal of one phase at this constant temperature. For water: H 2 O (solid) H 2 O (liquid) H ( fusion) = 6000 Joules / mole When subjected to an external pressure of one atmosphere, temperature is fixed at ZERO deg. Celsius H 2 O (liquid) H 2 O (gas) H ( vaporization) = 40.7 kj / mole When subjected to an external pressure of one atmosphere, temperature is fixed at 100 deg. Celsius TWO PHASES will always be present at boiling / condensation temperatures, and also at freezing / melting temperatures.

4 Steps in the solution of ex. 4: overall goal - bring both quantities of water to a common temperature and state, and evaluate net energy of the processes involved. In this example a common temperature of 50 degrees C and common state of liquid water will be targeted, An external pressure of one atmosphere is assumed. steps in the process: (1) COOL 40.0 g of gas from 125 to 100 degrees C. (sp.ht. gas = 2.02 J / g-deg) (2) CONDENSE 40.0 g of gas at 100 degrees C to liquid at 100 degrees C. ( H kj mole) ( vaporization) = / (3) COOL 40.0 g of liquid from 100 to 50 degrees C. (sp.ht. liquid = 4.18 J / g-deg) Obtain a SUBTOTAL of ENERGY involved in steps 1-3. (4) WARM 125 g of solid from 15 to ZERO degrees C. (sp.ht. solid = 2.03 J / g-deg (5) MELT 125 g solid to liquid at ZERO degrees C. ( H J mole) ( fusion) = 6000 / (6) WARM 125 g liquid from ZERO to 50 degrees C. (sp.ht. liquid = 4.18 J / g-deg) Obtain a SUBTOTAL of ENERGY involved in steps 4-6. (7) Determine NET ENERGY for steps 1 through 6. If NEGATIVE, then ADD net energy to total amount of liquid water at 50 degrees C, and calculate final temperature and state.

5 worked out steps: If POSITIVE, then subtract net energy from total amount of liquid water at 50 degrees C, and calculate final temperature and state. (1) q = ( J / g deg)( 40. 0g)( deg) q = 2020 Joules (sign follows from expression) (2)? q = kj g 18 g (sign must be furnished) CONDENSING involves REMOVAL OF HEAT, so sign of heat is NEGATIVE:? q = kj (3) q = ( 418. J / g deg)( 40. 0g)( deg)? q = 8,360 J (sign follows from expression) NET ENERGY STEPS 1 through 3 = 100,824 Joules (4) q = ( 2. 03J / g deg)( 125g)( 0 ( 15 deg))? q = J (sign follows from expression) 6.0 kj (5)?q = 125 g 18 g (sign must be furnished)

6 MELTING involves ADDITION OF HEAT, so sign of heat is POSITIVE:? q = + 41,667 J (6) WARM 125 grams liquid water from ZERO to 50 degrees C. q = (4.18J/g deg)(125g)(50 0 deg)? q = + 26,125 J (sign follows from expression) NET ENERGY STEPS 4 through 6 = + 71,598 Joules Condition of water at this stage: mass = 165 grams, state = liquid, Temperature = 50 deg.c (7) NET ENERGY STEPS 1 through 6 = -29,227 Joules Negative Energy means too much heat has been removed from the combined quantities of water to reach this condition. Consequently, it must be RETURNED to the total quantity of water, i.e., ADDED to it, so that NO NET ENERGY REMAINS. SO, ADDITION OF 29,227 Joules heat energy to 165 grams of liquid water at 50 degrees C will result in what final temperature? + 29, 227 J = (4.18 J/g deg)(165g)(t 50 deg) (final) T = 92 4 C Final State = LIQUID. ( final )