Question

clear explanion of the problem please

1. A 0.3-dm3 glass of water at 22°C is to be cooled with ice to 5°C. Determine how much ice needs to be added to the water, in grams, if the ice is at a) 0°C, b) -20°C. c) Also determine how much water would be needed if the cooling is to be done with cold water at 0°C. The melting temperature and the heat of fusion of ice at atmospheric pressure are 0°C and 333.67 kJ/kg, respectively, and the density of water is 1 kg/dm3. d) Using an appropriate software, investigate the effect of the initial temperature of the ice on the final mass required. Let the ice temperature vary from -26 to 0°C. Plot the mass of ice against the initial temperature of ice and discuss the results.

Answer 1

water is cooled from 22 C to 5 C using ice at 0 C, during this heat transfer process, the final equilibrium temp of both water in glass and ice will be 5 C.....now since the ice will melt and become water as its temperature drops from 0 C, Total heat transfered by ice will be fusion heat at 0 C and from 0 C to 5 C as water.

ii) If ice is at -20 C then three type of heat transfer takes place ....first from ice at -20 C to 0 C then fusion heat at 0 C then from 0 C to 5 C as water.

Sol 3 volume of glass 0.3x10 m = 0.3 dm – 22°C = 295B initial temp of water feriel tomb of water in glass (to be needed) – 52= 278K nlow density of water kg/m (at 22°c) - mass of water in glass weater =1000 3 3 V xs = 0.3x10 x 10 mg = 0.3 kg of water = Cpw specific heat of water 4.18 kJ/ky k to cool (i) ice at oc let mi = mass of ice needed ... heat lost ky water in glass water will be heat gained by ice <( 76-7,) heat lost by mg & Cow x = 0.3x 4.18 */295– 278) q = 21.318 kJ This amount of heat will be gained by ice q = (mex LiH). + mix Cow (278–273) 2. RJ / kg 1.21.318 LH af ice at o'c = 333.67 mi X 333-67 + mix4.18 (5) im; = 60017 gm ..mi 21.318 354.27

is mass of 5c is ice needed to cool water en glass from azt to 60.17 gm (ij) ue is at -20°C Specific heat of we = 2004 R 3 / ng K Latent heat of ice ato c = 33367 kJ/ky = CPC 24 Nlow the same amount of heat lost by weater in gloss will be gained by ice at -20°c id ::9= m; x 64 x (0-1-22] +fm;xly) + me x Cowl 5-0) (-e4 sao) .: 21.318 333•67 + (4:18 * s ) XS + mi mi = 53.92 gm - mi 21.318 395-37 nlow mass is 53.92 a gm of ice (at-26c) needed to coal 0.3 kg of water in glass from aac to sc c) rlow cold water at o°C heat gained by cold water = mw x Cow $(5-0) q 21.318 .. 1.02 Rg - mo X 4018 X5 mw= 20.9 0.3 kg of water in :. cold water (at oc) needed to cool glass from aac to 5c is 1.02 kg

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Now, the relationship between mas of ice and its initial temperature is given by

21.318= m(ice) X specific heat of Ice X (0+ T initial) + m(ice) X Latent heat

(-26 e T, co [ +9€€€ + '2 hoop - ) ?w = 31€•10 (17* ?uft (2-0) Се; سا q=

Since ice is taking heat and and its tempeerature is increasing. Assuming it is not allowed to go below the temperature 0 C of the ice. Then the variation of mass of ice required is given by plot as shown below

m (ice) (gms) 78 76 74 72 m (ice) gms 70 m (ice) (ems) 68 66 52 -30 -25 -20 -10 in 0 -15 ice initial temperature

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