The enthalpy change for converting 1.00 mol of ice at -60.0°C to water at 80.0°C is ___________$\text { kJ. }$ The specific heats of ice,water,and steam are $2.09 \mathrm { Jg } ^ { - 1 } { } ^ { \circ } \mathrm { C } ^ { - 1 }$ $4.18 \mathrm { Jg } ^ { - 1 } { } ^ { \circ } \mathrm { C } ^ { - 1 }$ and $1.84 \mathrm { Jg } ^ { - 1 } { } ^ { \circ } \mathrm { C } ^ { - 1 }$ respectively.For $\mathrm { H } _ { 2 }$ O, $\Delta _ { \text {fus } } H$ = 6.01 kJ mol^-1,and $\Delta _ { \text {vap } } H = 40.67 \mathrm {~kJ} \mathrm {~mol} ^ { - 1 }$ . A)9.77 B)14.29 C)8282 D)13.17 E)48.95

Question

Quizplus · Accepted Answer

The enthalpy change involves heating ice from -60°C to 0°C, melting ice to water, and then heating water to 80°C. Calculate each step:  1. Heating ice: $ q_1 = m \cdot c_{\text{ice}} \cdot \Delta T = 18 \, \text{g} \cdot 2.09 \, \text{J/g°C} \cdot 60°C = 2262 \, \text{J} $.  2. Melting ice: $ q_2 = \Delta H_{\text{fus}} = 6.01 \, \text{kJ/mol} = 6010 \, \text{J} $.  3. Heating water: $ q_3 = m \cdot c_{\text{water}} \cdot \Delta T = 18 \, \text{g} \cdot 4.18 \, \text{J/g°C} \cdot 80°C = 6019.2 \, \text{J} $.  Total $ q = q_1 + q_2 + q_3 = 2262 + 6010 + 6019.2 = 14291.2 \, \text{J} = 14.29 \, \text{kJ} $.

The Enthalpy Change for Converting 1 kJ. \text { kJ. } kJ. The Specific Heats of Ice,water,and Steam Are

The Enthalpy Change for Converting 1 $\text { kJ. }$ The Specific Heats of Ice,water,and Steam Are