Answer to Question #109109 in General Chemistry for Emme Ross

Question

If 6,880 Joules of heat are added to a 15.0-gram sample of solid ice initially at a temperature of -10.0 °C, calculate the final temperature of the resultant liquid water, assuming no heat is lost to the surroundings.

Answer 1

q = mass × C x ∆T

C - specific heat capacity

Heating of ice from -10 °C to 0°C

∆T = 0 - (-10) = 10°C

q = 15.0 g x 2.108 J/(g °C) x 10°C = 316.2 J

Fusing of ice into liquid water

q = m × ∆H_f

∆H_f - heat of fusion

q = 15.0 g x 334 J/g = 5010 J

Heating of water from 0°C to T final

q = 6880 - 316.2 - 5010 = 1553.8 J

∆T = q / (mass x C) = 1553.8 J / (15.0 g x 4.184 J/(g °C)) = 24.8°C

T_f = 24.8 - 0 = 24.8°C