Question

at 1 atm, how much energy is required to heat 79.0 g h₂o(s) at -10.0 °c to h₂o(g) at 143.0 °c? use the heat transfer constants found in this table.

Explanation:

Step1: Heat ice from - 10.0°C to 0.0°C

The specific - heat capacity of ice ($c_{ice}$) is used. The formula for heat transfer is $q = mc\Delta T$. The molar mass of $H_2O$ is $M = 18.02\ g/mol$, and the number of moles of $H_2O$, $n=\frac{m}{M}=\frac{79.0\ g}{18.02\ g/mol}\approx4.38\ mol$. The specific - heat capacity of ice $c_{ice}=2.09\ J/g\cdot^{\circ}C$, $\Delta T_1=0 - (- 10.0^{\circ}C)=10.0^{\circ}C$, $m = 79.0\ g$. So $q_1=mc_{ice}\Delta T_1=79.0\ g\times2.09\ J/g\cdot^{\circ}C\times10.0^{\circ}C = 1651.1\ J$.

Step2: Melt ice at 0.0°C

The heat of fusion of ice ($\Delta H_{fus}$) is used. $\Delta H_{fus}=6.01\ kJ/mol$. $q_2=n\Delta H_{fus}=4.38\ mol\times6.01\ kJ/mol = 26.32\ kJ=26320\ J$.

Step3: Heat water from 0.0°C to 100.0°C

The specific - heat capacity of water ($c_{water}$) is used. $c_{water}=4.184\ J/g\cdot^{\circ}C$, $\Delta T_3=100.0^{\circ}C - 0.0^{\circ}C = 100.0^{\circ}C$, $m = 79.0\ g$. So $q_3=mc_{water}\Delta T_3=79.0\ g\times4.184\ J/g\cdot^{\circ}C\times100.0^{\circ}C = 33053.6\ J$.

Step4: Vaporize water at 100.0°C

The heat of vaporization of water ($\Delta H_{vap}$) is used. $\Delta H_{vap}=40.7\ kJ/mol$. $q_4=n\Delta H_{vap}=4.38\ mol\times40.7\ kJ/mol = 178.37\ kJ=178370\ J$.

Step5: Heat steam from 100.0°C to 143.0°C

The specific - heat capacity of steam ($c_{steam}$) is used. $c_{steam}=2.01\ J/g\cdot^{\circ}C$, $\Delta T_5=143.0^{\circ}C - 100.0^{\circ}C = 43.0^{\circ}C$, $m = 79.0\ g$. So $q_5=mc_{steam}\Delta T_5=79.0\ g\times2.01\ J/g\cdot^{\circ}C\times43.0^{\circ}C = 6849.57\ J$.

Step6: Calculate the total energy

$q_{total}=q_1 + q_2+q_3+q_4+q_5$.
$q_{total}=1651.1\ J+26320\ J+33053.6\ J+178370\ J+6849.57\ J$
$q_{total}=246244.27\ J\approx246\ kJ$.

Answer:

$246\ kJ$