How many grams of ice at -28.9°C can be completely converted to a liquid at 24.4 ° C if the available heat for this process is 4.44x10^3kJ? For ice, use a specific heat of 2.01 J/(gc°C) and

Question

How many grams of ice at -28.9°C can be completely converted to a liquid at 24.4 ° C if the available heat for this process is 4.44x10^3kJ? For ice, use a specific heat of 2.01 J/(gc°C) and Delta H (fus)=6.01kJ/mol. Express your answer to 3 sig figs.

Answer 1

These problems are best solved using a step-wise approach.

First, we see that we have different units for ∆H and specific heat for ice, specific heat for liquid water (4.184 J/gº) and also g and moles, so we'll change the units of ∆Hfus and heat.

∆H_fus = heat to melt ice @ 0ºC = 6.01 kJ/mol x 1 mol/18 g = 0.333 kJ / g = 333 J/g

q = heat available = 4.44x10⁴ kJ = 4440 J

m = mass in grams =?

C = specific heat in J/gº

∆T = change in temperature in degrees

C_ice = 2.01 J/gº

C_H2O = 4.184 J/gº

heat needed to raise temp of ice from -28.9º to 0º = mC∆T = (m)(2.01 J/gº)(28.9º)

heat needed to melt the ice @ 0ºC = (m)(∆H_fus) = (m)(333 J/g)

heat needed to raise temp of liquid water from 0º to 24.4º = mC∆T = (m)(4.184 J/gº)(24.4º)

We know how much heat is available (4.44x10³ kJ = 4440 J) and so we can solve for the mass (m) by adding up all the above steps and setting it equal to the heat available.

4440 J = (m)(2.01 J/gº)(28.9º) + (m)(333 J/g) + (m)(4.184 J/gº)(24.4º)

4440 J = 58.09m + 333m + 102.1m

4440 J = 493.2 m

m = mass of ice = 9.00 g ice (3 sig. figs.)

please be sure to check the math