Question

Calculate the Heat of Fusion of Ice if you had 15 grams of warm water at 40.0oC in a Styrofoam cup and you added 4.27g of ice originally at -0.3oC and after the ice melted the combined liquid was 10.0oC and the specific heat of water is 1.00 cal/goC, what would be the calculated heat of fusion for water? Show all calculations in order. What is the percent error?

Answer 1

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg∙K), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

Heat lost by ice = Heat gained by water

Heat gained by water = 15 x 4.184 x (40-30) = 1882.8 Joules

Heat lost by ice = 4.27 x 2.06 x 0.3 + Heat of fusion of 4.27 gm ice + 4.27 x 4.184 x 10  

4.27 Heat of fusion + 181.295 =  1882.8

Heat of fusion calculated = 398.478 J/g∙C

Heat of fusion thoretical = 334.16 J g¯¹

Percentage error = (398.478 - 334.16) x 100 /334.16 = 19.24