
A couple sitting in a warm room on a winter day takes half kg of cheese sandwiches (an energy intake of \[8{\rm{ }}130{\rm{ }}\,kJ\] for both). The enthalpy of vaporisation of water is $40.65\, kJmol^{-1}$. Supposing that one of the energy is stored in body, mass (in g) of water would they need to perspire in order to maintain their original temperature is (nearest integer)
Answer
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Hint: Enthalpy which is denoted as \[H\] is the summation of the internal energy denoted \[U\] and the product of volume and pressure \[(PV)\] and solution is made by the equation \[H = U + PV\]. When a process occurs at constant pressure and volume, the heat evolved which is either released or absorbed, is equal to the difference in enthalpy.
Complete answer
The overall enthalpy of a system can not be defined directly, due to the fact the internal energy carries components which are not found or difficult to access, or basically not of interest in thermodynamics. In practical terms, a difference in enthalpy i.e, \[(\Delta H)\] is the expression for measurements and calculation at steady pressure, as it simplifies the outline of energy transferring. Enthalpy for chemical substances at steady pressure generally refer to preferred state most normally \[1bar\] \[(100kPa)\] pressure. Standard state of this does not strictly specify a temperature, however expressions for enthalpy usually reference the standard amount of a heat of formation at \[25^\circ C(298K)\]. For endothermic reaction processes, the differential change, \[(\Delta H)\] is a positive numeric value, and is negative value in exothermic reaction i.e, (heat-releasing) reactions.
Energy intake \[ = 8.130\,kJ\],
The enthalpy of vaporization for water is given by,
\[ = 40.65\,kJ/mol = \dfrac{{40.65\,kJ/mol}}{{18\,g/mol}} = 2.258\,kJ/g\]
Enthalpy of vaporization for water is \[2.258\,kJ/g\],
Thus, the mass (in gram) of water needed to perspire in order to maintain original body temperature ,
\[ = \dfrac{{8.310\,kJ}}{{2.258\,kJ/g}} = 3.60 \approx 4\, g\]
The mass (in gram) of water needed to perspire in order to maintain original body temperature is \[3.60 \approx 4\,g\]
Therefore the mass of water in grams is approx.4 grams.
Note:
The enthalpy of a perfect ideal gas is independent of its pressure numeric value, and it relies upon only on its specific temperature, which relates to the amount of its internal energy. Real gases at normal temperatures and pressures often carefully approximate this behavior, which simplifies the practical thermodynamic designation and analysis.
Complete answer
The overall enthalpy of a system can not be defined directly, due to the fact the internal energy carries components which are not found or difficult to access, or basically not of interest in thermodynamics. In practical terms, a difference in enthalpy i.e, \[(\Delta H)\] is the expression for measurements and calculation at steady pressure, as it simplifies the outline of energy transferring. Enthalpy for chemical substances at steady pressure generally refer to preferred state most normally \[1bar\] \[(100kPa)\] pressure. Standard state of this does not strictly specify a temperature, however expressions for enthalpy usually reference the standard amount of a heat of formation at \[25^\circ C(298K)\]. For endothermic reaction processes, the differential change, \[(\Delta H)\] is a positive numeric value, and is negative value in exothermic reaction i.e, (heat-releasing) reactions.
Energy intake \[ = 8.130\,kJ\],
The enthalpy of vaporization for water is given by,
\[ = 40.65\,kJ/mol = \dfrac{{40.65\,kJ/mol}}{{18\,g/mol}} = 2.258\,kJ/g\]
Enthalpy of vaporization for water is \[2.258\,kJ/g\],
Thus, the mass (in gram) of water needed to perspire in order to maintain original body temperature ,
\[ = \dfrac{{8.310\,kJ}}{{2.258\,kJ/g}} = 3.60 \approx 4\, g\]
The mass (in gram) of water needed to perspire in order to maintain original body temperature is \[3.60 \approx 4\,g\]
Therefore the mass of water in grams is approx.4 grams.
Note:
The enthalpy of a perfect ideal gas is independent of its pressure numeric value, and it relies upon only on its specific temperature, which relates to the amount of its internal energy. Real gases at normal temperatures and pressures often carefully approximate this behavior, which simplifies the practical thermodynamic designation and analysis.
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