Show all calculations and reasoning. How many ice cubes (at 0.0 degrees) would y
ID: 966265 • Letter: S
Question
Show all calculations and reasoning.How many ice cubes (at 0.0 degrees) would you have to add to a 100.0mL drink initialy at 25.0 degrees in order to lower its temperature to 4.0degrees? Each ice cube weights 5.00g, and the density of water is 1.00 g mL. Assume no heat loss. Fusion=6.02 kj /mol vapour=40.7 kj/mole Specific heat capacity of solid ice = 2.06 J /g C Spec. heat of liquid water = 4.184 J/g C Show all calculations and reasoning.
How many ice cubes (at 0.0 degrees) would you have to add to a 100.0mL drink initialy at 25.0 degrees in order to lower its temperature to 4.0degrees? Each ice cube weights 5.00g, and the density of water is 1.00 g mL. Assume no heat loss. Fusion=6.02 kj /mol vapour=40.7 kj/mole Specific heat capacity of solid ice = 2.06 J /g C Spec. heat of liquid water = 4.184 J/g C
How many ice cubes (at 0.0 degrees) would you have to add to a 100.0mL drink initialy at 25.0 degrees in order to lower its temperature to 4.0degrees? Each ice cube weights 5.00g, and the density of water is 1.00 g mL. Assume no heat loss. Fusion=6.02 kj /mol vapour=40.7 kj/mole Specific heat capacity of solid ice = 2.06 J /g C Spec. heat of liquid water = 4.184 J/g C
Explanation / Answer
Mass of drink = density * volume
= 1 g/mL * 100 mL
= 100 g
Heat given out by water/drink = m*C*delta T
= 100*4.184*(25-4)
= 8786.4 J
This heat must be absorbed by ice
Let total mass of ice be m' g
L = 6.02 KJ/mol = 6.02*10^3 J/mol = 6.02*10^3/18 J/g = 334.44 J/g
Heat absorbed = m'*L + m'*C'*delta T
8786.4 = m'*(334.44 + 2.06*(4-0))
8786.4 = m'*(334.44 + 2.06*4)
8786.4 = m'*(342.68)
m'= 25.64 g
mass of 1 cube = 5 g
so roughly one must use 5 cubes
Answer: 5 cubes