Question #109109
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.
Expert's answer
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 × ∆Hf
∆Hf - 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
Tf = 24.8 - 0 = 24.8°C
