In: Chemistry
Find the heat of fusion of hte ice. You will do this by immersing a 30mL ice cube into 150 mL of hot water
T ice = 0C
T water = 60C
T final = 43C
mass of cup: 24g
mass of ice: 28g
Mass of water: 130g
specific heat of liquid water = 4181 J/kg K
specific heat of ice = 2090 J/kg K
specific heat of aluminum (cup) = 900 J/kg K
First, identifty whats going to happen
the ice will transform from solid ice at 0ºC to liquid water at 0ºC
then, the liquid ice will go from 0ºC to Tfinal, which is 43ºC
for the water (hot water)
Water goes from 60ºC to 43ºC,
knowing this
Qice + Qcup = -Qwater
Qice + Qcup = mice*Latent heat + mice*Cwater*(Tf-Ti) + mcup*Calum*(Tf-Ti)
Qwater = mwater*Cwater*(Tf-Tw)
now,
mice*Latent heat + mice*Cwater*(Tf-Ti) + mcup*Calum*(Tf-Ti)= -mwater*Cwater*(Tf-Tw)
now, substitute all known data
28*LH + 28*4.184*(43-0) +24*0.90*(43-0)= -130*4.184*(60-43)
now, solve fot LH, latent heat of ice
28*LH + 28*4.184*(43-0) +24*0.90*(43-0)= -130*4.184*(60-43)
28*LH +5966.336 = - 9246.64
LH = (-9246.64 - 5966.336)/(28) = 543.320 J/g C approx