Question

What is the heat change in kJ associated with 55.59 g of ice at -5.00 ° C changing to liquid water at +5.00 °C?

c(H₂O) (liq) = 4.184 J/(g.K)

c(H₂O) (s) = 2.09 J/(g.K)

ΔH_fus(H₂O) = 6.02 kJ/mol

Answer 1

Answer 2

 Heating the ice from -5.00 °C to 0.00 °C

1. Use the specific heat capacity of ice:
   $q_1=m\cdot c_{\text{ice}}\cdot \mathrm{\Delta }T$
   
   $q_1=55.59\times 2.09\times 5.00=580.91\text{J}$
   $q_1=0.581\text{kJ}$

3. Melting the ice at 0.00 °C
   Use the heat of fusion:
   Moles of ice = $\frac{55.59\text{g}}{18.015\text{g/mol}}=3.085\text{mol}$
   $q_2=n\cdot \mathrm{\Delta }{H}_{\text{fus}}$
   $q_2=3.085\text{mol}\times 6.02\text{kJ/mol}=18.57\text{kJ}$

5. Heating the liquid water from 0.00 °C to 5.00 °C
   Use the specific heat capacity of liquid water:
   $q_3=m\cdot c_{\text{water}}\cdot \mathrm{\Delta }T$
   
   $q_3=55.59\times 4.184\times 5.00=1162.9\text{J}$
   $q_3=1.163\text{kJ}$

Total heat change:
$q_{\text{total}}=q_1+q_2+q_3$
$q_{\text{total}}=0.581\text{kJ}+18.57\text{kJ}+1.163\text{kJ}=20.31\text{kJ}$

Final Answer:
The total heat change is $\boxed{{\displaystyle 20.31\text{kJ}}}$