121. what quantity of energy does it take to convert 0.500 kg ice at - 20.0°c to steam at 250.0°c? specific…

Explanation:

Step1: Convert mass of ice to grams

$m = 0.500\ kg=500\ g$

Step2: Calculate heat to raise ice temperature to 0°C

$q_1 = mc\Delta T$, where $m = 500\ g$, $c = 2.03\ J/g\cdot^{\circ}C$, $\Delta T=20.0^{\circ}C$. So $q_1=500\ g\times2.03\ J/g\cdot^{\circ}C\times20.0^{\circ}C = 20300\ J$

Step3: Calculate heat for phase - change of ice to water

$q_2 = n\Delta H_{fus}$, molar mass of water $M = 18.02\ g/mol$, $n=\frac{m}{M}=\frac{500\ g}{18.02\ g/mol}\approx27.75\ mol$, $\Delta H_{fus}=6.02\ kJ/mol = 6020\ J/mol$. So $q_2=27.75\ mol\times6020\ J/mol = 166055\ J$

Step4: Calculate heat to raise water temperature to 100°C

$q_3 = mc\Delta T$, where $m = 500\ g$, $c = 4.18\ J/g\cdot^{\circ}C$, $\Delta T = 100^{\circ}C$. So $q_3=500\ g\times4.18\ J/g\cdot^{\circ}C\times100^{\circ}C=209000\ J$

Step5: Calculate heat for phase - change of water to steam

$q_4 = n\Delta H_{vap}$, $n\approx27.75\ mol$, $\Delta H_{vap}=40.7\ kJ/mol = 40700\ J/mol$. So $q_4=27.75\ mol\times40700\ J/mol = 1129425\ J$

Step6: Calculate heat to raise steam temperature to 250°C

$q_5 = mc\Delta T$, where $m = 500\ g$, $c = 2.02\ J/g\cdot^{\circ}C$, $\Delta T=150^{\circ}C$. So $q_5=500\ g\times2.02\ J/g\cdot^{\circ}C\times150^{\circ}C = 151500\ J$

Step7: Calculate total heat

$Q=q_1 + q_2+q_3+q_4+q_5$ $Q=20300\ J+166055\ J+209000\ J+1129425\ J+151500\ J = 1676280\ J=1676.28\ kJ$

Answer:

$1676.28\ kJ$