Question

Question 15 (1 point) Steam at 100° C is added to ice at 0° C. Find the amount of ice melted and the final temperature when the mass of steam is 7.00 g and the mass of ice is 50.0 g. 50.0 g, 11.7°C 40.0 g, 8.80° C 30.0 g, 8.80° C 34.7 g 0° C 50.0 g, 8.80° C

Answer 1

Energy needed to melt 50g of at ice is max Mass 2 , mm; - in 4 of 50 = 0.05 ko gog 0,054 = 0.05 X 3 -3 3x10.5 4 = Latent heard = 16650 of fusion of in 3/19.0c 3.33x105 ou

Now the heat needed for Fg of rapour to condense is mg = mass of steam 6 = 0:00.7 x 2.26X10 = 15820 J 6 -2.26X10 Qz = mg xlu =7g 79 z o.oof 07 by Lo=Latent heat of condensation' 8J/kg.de See that ligye. That means comes from condensation of sapoune is not enough to melt the all in, so the final temperature is oºc we have to calculate how much in is melted. Here we can heart Now

Since the final temperature is o'c, so all the infrunt will come to ooc. too°c So hent needed fore 7g unpouse to come from 100°c is Q ms XC W X AT 2 Co = specific heart of watere -0.007 X 4186 X100 2930:23 Total heat Quest and 3 15820 - 293002 12889.8] So, if en is the ine melled, then Q mx4 'heart ms 12889.8 3.33x 10 5 to 0.0387 38.78 34177 Nearest answere is 34.7 So right and here is 4the option.

ideal gas an iso there Mwam Vi The work done during mal process is giren by W = nRT ln V Ts constant temperature of gas Final volume of gas of gas R= Universal gas PV = nRT V ; = Initial vo lume of g gas No M of moles gas chustand Wilm Vi क ) MRT W/ART V: V. t Vre VA : vf ew/PEVA -3 -W/MRT Ve Bcm/(1013 X10x 46X163) = 46810 xe = 210244 L I since the work is done on gas, megelive Vi = 21.244 L 21.244x10-3 so we will be

b ) 1 Again hRT. PV. f vf vf Т. MR 5 3 1.013 X10 3x10 X 46x10 2 x 8.314 2 280.24K Nearest option 7.24°C option 6.86°c right option is second option. So