Question

Calculate the amount of heat absorbed by 10.0 grams of ice at -15.0°C in converting it to liquid water at 50.0°C. (Sp. heat of H2O(s) = 2.09 J/g•°C, Sp. heat of H2O( ) = 4.18 J/g•°C, heat of fusion of H2O(s) = 333 J/g)
a. 5.73  103 J
b. 6.76  102 J
c. 1.70  102 J
d. 2.83  103 J
e. 3.29  103 J

Answer 1

The energy associated with temperature change for ice is;

$\Delta H=m\times C\times \Delta T$

$\Delta H$ $=(10g)($ $2.09J\over g.°C$ $)$ $(15°C)$

$=313.5J$

The energy of fusion in melting$10g$ of ice requires;

$10gH_2O\times$ $1mol H_2O\over 18.02gH_2O$ $=0.555mol H_2O$

$\Delta H_{fus}=6.0kJ/mol$ (Molar heat of fusion in melting)

$\therefore$ $\Delta H$ $=0.555molH_2O$ $\times$ $6.0kJ\over 1mol H_2O$ $=3.33 kJ=3330J$

Energy required to change the the temperature of liquid water is;

$\Delta H=10g$ $\times$ $($ $4.18J\over g.°C$ $)$ $(50°C)$ $=2090J$

The energy required for the entire change is;

$2090J+3330J+313.5J=5733.5J$

$=5.73\times 10^{3}J$

The correct answer is $A$