6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)​

Respuesta :

Taking into account the definition of calorimetry, 0.0185 moles of water are required.

Calorimetry

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

Mass of water required

In this case, you know:

  • Heat= 92.048 kJ
  • Mass of water = ?
  • Initial temperature of water= 34 ºC
  • Final temperature of water= 100 ºC
  • Specific heat of water = 4.186 [tex]\frac{J}{gC}[/tex]

Replacing in the expression to calculate heat exchanges:

92.048 kJ = 4.186 [tex]\frac{J}{gC}[/tex]× m× (100 °C -34 °C)

92.048 kJ = 4.186 [tex]\frac{J}{gC}[/tex]× m× 66 °C

m= 92.048 kJ ÷ (4.186 [tex]\frac{J}{gC}[/tex]× 66 °C)

m= 0.333 grams

Moles of water required

Being the molar mass of water 18 [tex]\frac{g}{mole}[/tex], that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:

[tex]amount of moles=0.333 gramsx\frac{1 mole}{18 grams}[/tex]

amount of moles= 0.0185 moles

Finally, 0.0185 moles of water are required.

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