Suppose 400.0g of water at 30.0°C is poured over a 69.0-g cube of ice with a tem
ID: 1063959 • Letter: S
Question
Suppose 400.0g of water at 30.0°C is poured over a 69.0-g cube of ice with a temperature of -5.00°C. If all the ice melts, what is the final temperature of the water? if all the ice does not melt, how much ice remains when the water-ice mixture reaches equilibrium?
May I ask how would you know if the ice melts all of the way or not? Please explain for me lol :)
400.0 g of water at is poured over a cube temperature of 5.00 c. If all of the ice melts, what is the final temperature of the water? if all of the ice does not melt, how much ice remains when the water-ice mixture reaches equilibrium?Explanation / Answer
from -5 deg.c to 0 deg.c the ice needs to be supplied with sensible heat and this is known as sensible heat
sensible heat = mass of ice* specific heat of ice* temperature differece = 69*2.05*(5-(-0)= 707.25 joules
Assuming complete ice melts, ice needs to be supplied with latent heat ( whose value is 335 J/g)
Heat required for melting of ice = 69*335 joules =23115 joules
Total heat to be supplied for melting of ice from -5 deg.c to 0 deg.c= 23115+707.25 joules=23822.25 joules.
It has to be seen whether this much heat can be provided by brining the temperature of 400 gm of water
Sensible heat of 400 gm of water from 30 deg.c to 0 deg.c = 400*4.184*(30-0)=50208 joules.Since this heat is much more than the heat required for complete melting.
let the final temperatue be T
heance heat lost by 400 gm of water =heat gained by water ( sensible heat ( from -5 to 0 deg.c+ latent heat for converring ice at 0 deg. c to liquid water at 0 deg.c + sensible heat of liquid water from 0 to T)
hence 400*4.184*(30-T)= 23822.25+69*4.184*(T-0)
1673.6*(30-T)= 23822.25+289*(T-0)
50208- 1673.6T= 23822.25+289T
50208-23822.25= T*(289+1673.6)
T= 13.44 deg.c
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.