Question

3. What is the final temperature and physical state of H2O when Add 250g of liquid at 85C to 80 g of ice at -15C? Helpful Information: Cs(ice) = 2.1 J/g·°C, ΔHfus(water) = 6.02 kJ/mol, Cs(water) = 4.184 J/g·°C, ΔHvap(water) = 40.7 kJ/mol,Cs(steam) = 2.0 J/g·°C.

Answer 1

Heat lost by liquid water =250*4.18*(85-T)=88825-1045T

1. from-15 deg.c to 0 deg.c, sensible heat of ice =mass of ice* specifc heat of ice* temperature difference= 80*2*15 2400 Joules

2. at 0 deg.c it gains heat of fusion ,heat transferred due to heat of fusion=(80/18)*6.02*1000 joules= 26764.44joules

Since total heat from 1 and 2 is only 2400+26764.44=29164,44    which is compared to 88825-1045T is less, liquid water can also take up sensible heat and reaches some temperature T.

Sensilbe heat= 80*4.18*(T-0)

So total heat= 29164.4+334.4T= 88825-1045T

T*(1045+334.4)= 88825-29164=59601

T= 43.25 deg.c,

So the temperature is 43.25 deg,c and it is liquid water.