Question

How much heat (in $ {{kJ}} $ ) is needed to convert $ 866\;{{g}} $ of ice at $ - {10^\circ }{{C}} $ to steam at $ {126^\circ }{{C}}? $ (The specific heats of ice and steam are $ 2.03\;{{J}}/{{g}}{.^\circ }{{C}} $ and $ {{1}}{{.99J}}/{{g}}{.^\circ }{{C}}, $ respectively. )

Answer 1

Hint: A substance's heat capacity is the amount of heat required to increase the temperature of the entire substance by one degree. If the mass of the substance is united, the specific heat capacity or the specific heat is called the heat capacity.

Formula used:
We will use the following formulas to get the solution to the above problem
 $ q = mc\Delta T $
 $ q = m{H_{{{fusion }}}} $
 $ q = m{H_{{{vaporization }}}} $
Where
 $ q $ is the Heat/energy
 $ m $ is the Mass
 $ c $ is the Specific heat
 $ \Delta T $ is the change in temperature
 $ {H_{{{fusion }}}} $ is the energy required to melt/freeze one gram of a substance
 $ {H_{{{vaporization }}}} $ is the energy required to vaporize/condense one gram of a substance.

Complete step by step solution:
Let us suppose the heat required to raise the ice to its melting point $ \left( {{0^\circ }{{C}}} \right) $ be $ {q_1} $
Let us suppose the heat required to melt all of given ice be $ {q_2} $
Let us suppose the heat required to raise the temperature of the water to its boiling point $ \left( {{{100}^\circ }{{C}}} \right) $ be $ {q_3} $
Let us suppose the heat required to vaporize all the water be $ {q_4} $
And the heat required to raise the temperature of the steam to $ {126^\circ }{{C}} $ be $ {q_5} $
Now, we will write all the equations with the given information
 $ {q_1} = 866{{g}} \times \dfrac{{2.03{{J}}}}{{{{{g}}^{{^\circ }}}{{C}}}} \times {10^\circ }{{C}} = 17579.8{{J}} $
 $ {q_2} = 866{{g}} \times \dfrac{{334{{J}}}}{{{g}}} = 289244{{J}} $
 $ {q_3} = 866{{g}} \times \dfrac{{4.18{{J}}}}{{{{{g}}^{{^\circ }}}{{C}}}} \times {100^\circ }{{C}} = 361988{{J}} $
 $ {q_4} = 866{{g}} \times \dfrac{{2260{{J}}}}{{{g}}} = 1957160{{J}} $
 $ {q_5} = 866{{g}} \times \dfrac{{1.99{{J}}}}{{{{{g}}^{{^\circ }}}{{C}}}} \times {26^\circ }{{C}} = 44806.84{{J}} $
The total heat required to convert ice to steam at the required conditions will be the sum of all the above values
That is,
 $ {q_1} + {q_2} + {q_3} + {q_4} + {q_5} $
Let us now substitute the values to get
 $ 17579.8 + 289244 + 361988 + 1957160 + 44806.84 $
Upon solving, we get
 $ q = 2670778.64{{ J}} $
Or we can rewrite it as
 $ \therefore q = 2670.77{{ kJ}} $ .

Note:
Heat, often called thermal energy, is a form of energy. Energy can be transformed from one form to, but it can neither be generated nor destroyed; energy is conserved instead.