Question

GOAL Solve a problem involving heat transfer and a phase change from solid to liquid. PROBLEM At a party, 6.00 kg of ice at -5.00°C is added to a cooler holding 30 liters of water at 20.0°C. What is the temperature of the water when it comes to equilibrium? STRATEGY In this problem, its best to make a table. With the addition of thermal energy Dice the ice will warm to 0°C, then melt at 0°C with the addition of energy melt. Next, the melted ice will warm to some final temperature T by absorbing energy ice-water, obtained from the energy change of the original liquid water, water. By conservation of energy, these quantities must sum to zero. SOLUTION Calculate the mass of liquid water. mwater - Pwater = (1.00 x 103 kg/m3)(30.0L)(1.00 m3 / 1.00 x 103 L) - 30.0 kg (1) Dice + Omelt + Vice-water + Q water = 0 Write the equation of thermal equilibrium Construct a comprehensive table. (J/kg. "C) 2090 0 Dice melt Dice-water m (kg) 6.00 6.00 6.00 30.0 L (J/kg) Ty 0 3.33 X 1050 - T T T; -5.00 0 0 20.0 Expression iceice(T; -T;) micely icewat(T-T) wat wat(T-T) 4190 4190 Owater Substitute all quantities in the second through sixth columns into the last column and sum, which is the evaluation of Equation (1), and solve for T. 6.27 x 104 ] +2.00 x 106) + (2.51 x 104 1/°C)T-0°C) +(1.26 x 105 3/°C)(T - 20.0°C) = 0 T = 3.03°C

Answer 1

1 * from thermal equilibrium :- Qice & Sonett + Qice-water t. Qwator zo mice Cicero--148)} + mice * Lf + miceCwatert (16-0) E t mwater * Cwater* (TF-20) A hor mice = mars of ice = 4 kg mwaters Density * volume = Lood2 30kg Cice = 2090 J/kg-ic , Cwater = 4.100 Ilgoc Lf= 3.33x105 kg MADE 4 * 2090*11.8 +4* 3.33% 11.8 + 4* 3. 33x105 + 4*4190x Tf . a + 30% 4190*C 16-20170 Hency final equilibrium temp., Tf = 7.604°C] Amy . from thermal equilibriung :- Qice Camelt + Picewaton + water = 0 micex Cicex (0-(-1118))mice tlf + micex Cicercoatex*(TF-0) +mwater* Cwater * (T-20) o mice # 2090 x 11.8 + micet 3.33x105 + micex 4130x10. + (1800*1. 29x102) * 4190* (20-2020 a man of ice, mice 135 295.62