a) A 60.00 g piece of zinc, initially at 400.degree C, is dropped into 255 g of
ID: 525673 • Letter: A
Question
a) A 60.00 g piece of zinc, initially at 400.degree C, is dropped into 255 g of ice at-10.degree C. Calculate the final temperature of the system and identify the phases of water present. s_water, gas = 2.01 J/g degree C s_water, liq = 4.19 J/g degree C s_ice = 2.10 J/g degree C Delta H_vap = 2270 J/g Delta H_fus = 333 J/g s_zinc, solid = 0.388 J/g degree C MP_zinc = 419 degree C b) If the zinc were at 500. degree C instead of 400.degree C what additional information would you need? c) at weight of zinc at 400.degree C would be needed to heat the water to 125 degree C?Explanation / Answer
ice from -10 deg.c to 0 deg.c requires sensible heat and this is given by =mass* specific heat* temperature difference= 255*2.1*10= 5355 joules
at 0 deg.c , ice needs to be supplied with latent heat of fusion and this heat= mass* latent heat of fusion =255*333= 84195 Joules
zinc at 400 deg.c if cools down to 0 deg.c, heat need to be removed= mass* specific heat* temperature difference= 60*0.388*(400-)= 6208. This heat is adequate to convert ice at -10deg.c to ice at 0 deg.c
additional heat available =6208-5355= 853 Joules
this melts= 853/333 gm of water= 2.6 gm of water
since at equilibrium temperature of 0 deg.c, there will be two phase mixture of ice and water and zinc.
quantities ( gm ) ice = 255-2.6= 252.4, water= 2.6 and zinc= 60 gm
2. since zinc melts at 419 deg.c and the temperature given is 500 deg.c, one has to specify the specific heat of liquid zinc and latent heat of fusion of zinc.
3. ice requires 5355 joules to become ice at 0 deg.c and 84195 joules to become liquid at 0 deg.c
Additionally,
one needs to rise the temperature of water from 0 deg.c to 100 dec by application of sensible heat and this is given by= mass* specific heat* temperature difference= 255* 4.184*100 = 106692 joules. At 100 deg.c, one has to supply latent heat of vaporization and this heat= mass* latent heat of vaporization = 255*2270 = 578850 joules. From 100 deg.c to 125 deg.c, it is super heat of vapor and this = mass* specific heat* temperature difference= 255*2.01*25= 12813.75
total heat to be supplied is sum of sensible heat ( -10 deg.c to 0 deg.c) of ice + heat of fusion for melting ice at 0 deg.c+ sensible heat for rising the temperature of water from 0 deg.c to 100 deg.c + heat of vaporization to convert liquid water into vapor at 100 deg.c + super heat for converting vapor from 100 deg.c to 125 deg.c
=5355+84915+106692+578850+12813.75=788625.8 joules
all this heat has to come from loss of sensible heat of zinc at 400 deg.c
hence m* specific heat of zinc* temperature difference ( zinc ultimately has to reach equilibrium temperature of 125 deg.c )
hence m* 0.388*(400-125)= 788625.8
m= mass of zinc= 7391 gm of zinc = 7.391 Kg of zinc
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.